MBBS and BDS entrance Chemistry: NEET-UG questions on Nepal's MBBS syllabus: Physical Chemistry

2016 · NEET-UG (India) · 4 marks

The rate of a first-order reaction is $0.04\ \mathrm{mol\ L^{-1}\ s^{-1}}$ at 10 seconds and $0.03\ \mathrm{mol\ L^{-1}\ s^{-1}}$ at 20 seconds after initiation of the reaction. The half-life period of the reaction is: (a) 24.1 s (b) 34.1 s (c) 44.1 s (d) 54.1 s

Answer

(a) 24.1 s

For a first-order reaction rate $\propto$ concentration, so $\dfrac{r_1}{r_2}=\dfrac{0.04}{0.03}=e^{k(20-10)}$ . $k=\dfrac{2.303}{10}\log\dfrac{4}{3}=0.0288\ \mathrm{s^{-1}}$ , so $t_{1/2}=\dfrac{0.693}{0.0288}\approx 24.1$ s.

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2016 · NEET-UG (India) · 4 marks

$\mathrm{MY}$ and $\mathrm{NY_3}$ , two nearly insoluble salts, have the same $K_{sp}$ values of $6.2\times10^{-13}$ at room temperature. Which statement would be true in regard to $\mathrm{MY}$ and $\mathrm{NY_3}$ ? (a) The molar solubilities of MY and $\mathrm{NY_3}$ in water are identical. (b) The molar solubility of MY in water is less than that of $\mathrm{NY_3}$ . (c) The salts MY and $\mathrm{NY_3}$ are more soluble in 0.5 M KY than in pure water. (d) The addition of the salt of KY to solution of MY and $\mathrm{NY_3}$ will have no effect on their solubilities.

Answer

(b) The molar solubility of MY in water is less than that of $\mathrm{NY_3}$ .

For MY, $K_{sp}=s^2$ , so $s=\sqrt{6.2\times10^{-13}}\approx7.9\times10^{-7}$ M. For $\mathrm{NY_3}$ , $K_{sp}=27s^4$ , so $s=(6.2\times10^{-13}/27)^{1/4}\approx3.9\times10^{-4}$ M. Thus MY is less soluble. Adding KY (common ion Y $^-$ ) lowers the solubility of both, so (c) and (d) are wrong.

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2016 · NEET-UG (India) · 4 marks

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is: (a) $\Delta H<0$ and $\Delta S=0$ (b) $\Delta H>0$ and $\Delta S<0$ (c) $\Delta H<0$ and $\Delta S>0$ (d) $\Delta H<0$ and $\Delta S<0$

Answer

(c) $\Delta H<0$ and $\Delta S>0$

Spontaneity requires $\Delta G=\Delta H-T\Delta S<0$ . If $\Delta H$ is negative and $\Delta S$ positive, then $-T\Delta S$ is also negative at every temperature, so $\Delta G<0$ always. The other combinations either have $\Delta G$ depending on temperature or are non-spontaneous.

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2016 · NEET-UG (India) · 4 marks

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape? (a) 1/8 (b) 1/4 (c) 3/8 (d) 1/2

Answer

(a) 1/8

By Graham's law, $\dfrac{r_{H_2}}{r_{O_2}}=\sqrt{\dfrac{32}{2}}=4$ . In the same time, if half the hydrogen (0.5 of its moles) escapes, the oxygen escaping is $0.5/4=1/8$ of its moles. Since the two gases start with equal moles, the fraction of oxygen escaped is $1/8$ .

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2016 · NEET-UG (India) · 4 marks

Consider the following liquid-vapour equilibrium: Liquid $\rightleftharpoons$ Vapour. Which of the following relations is correct? (a) $\dfrac{d\ln G}{dT^2}=\dfrac{\Delta H_v}{RT^2}$ (b) $\dfrac{d\ln P}{dT}=\dfrac{-\Delta H_v}{RT}$ (c) $\dfrac{d\ln P}{dT^2}=\dfrac{-\Delta H_v}{T^2}$ (d) $\dfrac{d\ln P}{dT}=\dfrac{\Delta H_v}{RT^2}$

Answer

(d) $\dfrac{d\ln P}{dT}=\dfrac{\Delta H_v}{RT^2}$

This is the Clausius–Clapeyron equation for the liquid–vapour equilibrium, taking the vapour as ideal and neglecting liquid volume. It shows vapour pressure increases with temperature, so $d\ln P/dT$ must be positive; options (b) and (c) carry a wrong negative sign, and (a) uses $G$ instead of $P$ .

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2016 · NEET-UG (India) · 4 marks

The pressure of $\mathrm{H_2}$ required to make the potential of $\mathrm{H_2}$ -electrode zero in pure water at 298 K is (a) $10^{-14}$ atm (b) $10^{-12}$ atm (c) $10^{-10}$ atm (d) $10^{-4}$ atm

Answer

(a) $10^{-14}$ atm

The half-reaction is $\mathrm{2H^++2e^-\rightarrow H_2}$ , so $E=-\dfrac{0.059}{2}\log\dfrac{P_{H_2}}{[H^+]^2}$ . For $E=0$ , $P_{H_2}=[H^+]^2$ . In pure water $[H^+]=10^{-7}$ M, so $P_{H_2}=10^{-14}$ atm.

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2016 · NEET-UG (India) · 4 marks

The addition of a catalyst during a chemical reaction alters which of the following quantities? (a) Entropy (b) Internal energy (c) Enthalpy (d) Activation energy

Answer

(d) Activation energy

A catalyst provides an alternative pathway with lower activation energy, which speeds up the reaction. It does not change state functions such as entropy, internal energy or enthalpy, since initial and final states are the same.

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2016 · NEET-UG (India) · 4 marks

Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25 °C. (Given, vapour pressure data at 25 °C: benzene = 12.8 kPa, toluene = 3.85 kPa) (a) The vapour will contain a higher percentage of benzene. (b) The vapour will contain a higher percentage of toluene. (c) The vapour will contain equal amounts of benzene and toluene. (d) Not enough information is given to make a prediction.

Answer

(a) The vapour will contain a higher percentage of benzene.

In an ideal 1:1 mixture each mole fraction is 0.5, so partial pressures are $p_B=0.5\times12.8=6.4$ kPa and $p_T=0.5\times3.85=1.925$ kPa. Vapour mole fractions are proportional to partial pressures, so benzene (the more volatile component) is about 77% of the vapour. The data given is enough to decide.

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2016 · NEET-UG (India) · 4 marks

At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_b=0.52$ , the boiling point of this solution will be (a) 101 °C (b) 100 °C (c) 102 °C (d) 103 °C

Answer

(a) 101 °C

Relative lowering of vapour pressure: $\dfrac{760-732}{760}=\dfrac{n}{n+N}$ with $N=100/18=5.56$ mol, giving $n\approx0.21$ mol and molality $\approx2.1$ m. $\Delta T_b=K_bm=0.52\times2.1\approx1.1$ K, so the boiling point is about 101 °C.

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2016 · NEET-UG (India) · 4 marks

The molar conductivity of a $0.5\ \mathrm{mol/dm^3}$ solution of $\mathrm{AgNO_3}$ with electrolytic conductivity of $5.76\times10^{-3}\ \mathrm{S\ cm^{-1}}$ at 298 K is (a) $2.88\ \mathrm{S\ cm^2/mol}$ (b) $11.52\ \mathrm{S\ cm^2/mol}$ (c) $0.086\ \mathrm{S\ cm^2/mol}$ (d) $28.8\ \mathrm{S\ cm^2/mol}$

Answer

(b) $11.52\ \mathrm{S\ cm^2/mol}$

$\Lambda_m=\dfrac{\kappa\times1000}{M}$ with $M$ in mol/L. Here $0.5\ \mathrm{mol/dm^3}=0.5$ M. $\Lambda_m=\dfrac{5.76\times10^{-3}\times1000}{0.5}=11.52\ \mathrm{S\ cm^2\ mol^{-1}}$ .

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2016 · NEET-UG (India) · 4 marks

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is (a) 55 minutes (b) 110 minutes (c) 220 minutes (d) 330 minutes

Answer

(b) 110 minutes

At the anode $\mathrm{2Cl^-\rightarrow Cl_2+2e^-}$ , so 0.10 mol of $\mathrm{Cl_2}$ needs 0.20 mol of electrons: $Q=0.20\times96500=19300$ C. $t=Q/I=19300/3\approx6433$ s $\approx107$ min, closest to 110 minutes.

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2016 · NEET-UG (India) · 4 marks

For a sample of perfect gas when its pressure is changed isothermally from $p_i$ to $p_f$ , the entropy change is given by (a) $\Delta S=nR\ln\left(\dfrac{p_f}{p_i}\right)$ (b) $\Delta S=nR\ln\left(\dfrac{p_i}{p_f}\right)$ (c) $\Delta S=nRT\ln\left(\dfrac{p_f}{p_i}\right)$ (d) $\Delta S=RT\ln\left(\dfrac{p_i}{p_f}\right)$

Answer

(b) $\Delta S=nR\ln\left(\dfrac{p_i}{p_f}\right)$

For an isothermal change of a perfect gas, $\Delta S=nR\ln(V_f/V_i)$ and $V\propto1/p$ , so $\Delta S=nR\ln(p_i/p_f)$ . Entropy falls when pressure rises. Options with extra $T$ factors have the wrong units, and (a) has the sign inverted.

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2016 · NEET-UG (India) · 4 marks

The van't Hoff factor ( $i$ ) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (a) 0 (b) 1 (c) 2 (d) 3

Answer

(d) 3

$\mathrm{Ba(OH)_2}$ is a strong electrolyte that dissociates completely into three ions: $\mathrm{Ba(OH)_2\rightarrow Ba^{2+}+2OH^-}$ . In dilute solution the van't Hoff factor is therefore $i=3$ .

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2016 · NEET-UG (India) · 4 marks

The percentage of pyridine ( $\mathrm{C_5H_5N}$ ) that forms pyridinium ion ( $\mathrm{C_5H_5N^+H}$ ) in a $0.10\ M$ aqueous pyridine solution ( $K_b$ for $\mathrm{C_5H_5N}=1.7\times10^{-9}$ ) is (a) 0.0060% (b) 0.013% (c) 0.77% (d) 1.6%

Answer

(b) 0.013%

Pyridine is a weak base, so $\alpha=\sqrt{K_b/C}=\sqrt{1.7\times10^{-9}/0.10}=\sqrt{1.7\times10^{-8}}\approx1.3\times10^{-4}$ . As a percentage this is $1.3\times10^{-2}\%=0.013\%$ . Option (c) 0.77% would need a much larger $K_b$ (about $10^{-5}$ ).

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2016 · NEET-UG (India) · 4 marks

If the $E^\circ_{cell}$ for a given reaction has a negative value, which of the following gives the correct relationships for the values of $\Delta G^\circ$ and $K_{eq}$ ? (a) $\Delta G^\circ>0;\ K_{eq}<1$ (b) $\Delta G^\circ>0;\ K_{eq}>1$ (c) $\Delta G^\circ<0;\ K_{eq}>1$ (d) $\Delta G^\circ<0;\ K_{eq}<1$

Answer

(a) $\Delta G^\circ>0;\ K_{eq}<1$

$\Delta G^\circ=-nFE^\circ_{cell}$ , so a negative $E^\circ_{cell}$ gives $\Delta G^\circ>0$ . Since $\Delta G^\circ=-RT\ln K_{eq}$ , a positive $\Delta G^\circ$ means $\ln K_{eq}<0$ , i.e. $K_{eq}<1$ : the reaction is non-spontaneous as written and favours reactants.

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2016 · NEET-UG (India) · 4 marks

Which one of the following is incorrect for an ideal solution? (a) $\Delta H_{mix}=0$ (b) $\Delta U_{mix}=0$ (c) $\Delta P=P_{obs}-P_{calculated\ by\ Raoult's\ law}=0$ (d) $\Delta G_{mix}=0$

Answer

(d) $\Delta G_{mix}=0$

For an ideal solution $\Delta H_{mix}=0$ , $\Delta V_{mix}=0$ (so $\Delta U_{mix}=0$ ) and the vapour pressure obeys Raoult's law exactly, so $\Delta P=0$ . Mixing is spontaneous because entropy increases, so $\Delta G_{mix}=-T\Delta S_{mix}<0$ , not zero.

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2016 · NEET-UG (India) · 4 marks

The solubility of AgCl(s) with solubility product $1.6\times10^{-10}$ in $0.1\ M$ NaCl solution would be (a) $1.26\times10^{-5}\ M$ (b) $1.6\times10^{-9}\ M$ (c) $1.6\times10^{-11}\ M$ (d) zero

Answer

(b) $1.6\times10^{-9}\ M$

In 0.1 M NaCl, $[\mathrm{Cl^-}]\approx0.1$ M from the common ion, so $s=K_{sp}/[\mathrm{Cl^-}]=1.6\times10^{-10}/0.1=1.6\times10^{-9}$ M. Option (a), $1.26\times10^{-5}$ M ( $\sqrt{K_{sp}}$ ), is the solubility in pure water; the common ion suppresses it.

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2016 · NEET-UG (India) · 4 marks

The number of electrons delivered at the cathode during electrolysis by 1 ampere in 60 seconds is (charge on electron $=1.60\times10^{-19}\ \mathrm{C}$ ) (a) $6\times10^{23}$ (b) $6\times10^{20}$ (c) $3.75\times10^{20}$ (d) $7.48\times10^{23}$

Answer

(c) $3.75\times10^{20}$

Charge passed $Q=It=1\times60=60$ C. Number of electrons $=Q/e=60/(1.60\times10^{-19})=3.75\times10^{20}$ .

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2016 · NEET-UG (India) · 4 marks

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because (a) zinc is lighter than iron (b) zinc has lower melting point than iron (c) zinc has lower negative electrode potential than iron (d) zinc has higher negative electrode potential than iron

Answer

(d) zinc has higher negative electrode potential than iron

Zn ( $E^\circ=-0.76$ V) is more electropositive than Fe ( $-0.44$ V), so zinc is oxidised in preference and protects the iron sacrificially. Iron cannot protect zinc coating in reverse because it is the less reactive metal. Melting point or weight is irrelevant.

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2015 · NEET-UG (India) · 4 marks

A gas such as carbon monoxide would be most likely to obey the ideal gas law at: (a) low temperatures and low pressures. (b) high temperatures and low pressures. (c) low temperatures and high pressures. (d) high temperatures and high pressures.

Answer

(b) high temperatures and low pressures.

Real gases approach ideal behaviour when intermolecular attractions and molecular volume are negligible: high temperature (large kinetic energy) and low pressure (large volume, molecules far apart).

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2015 · NEET-UG (India) · 4 marks

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? (a) 1.04 (b) 12.65 (c) 2.0 (d) 7.0

Answer

(b) 12.65

Equal volumes: excess $\mathrm{OH^-}$ moles $\propto 0.1-0.01=0.09$ in twice the volume, so $[\mathrm{OH^-}]=0.045$ M. $\mathrm{pOH}=-\log0.045=1.35$ , so $\mathrm{pH}=14-1.35=12.65$ .

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2015 · NEET-UG (India) · 4 marks

Which one of the following pairs of solution is not an acidic buffer? (a) $\mathrm{H_3PO_4}$ and $\mathrm{Na_3PO_4}$ (b) $\mathrm{HClO_4}$ and $\mathrm{NaClO_4}$ (c) $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COONa}$ (d) $\mathrm{H_2CO_3}$ and $\mathrm{Na_2CO_3}$

Answer

(b) $\mathrm{HClO_4}$ and $\mathrm{NaClO_4}$

An acidic buffer needs a weak acid and its salt with a strong base. $\mathrm{HClO_4}$ is a strong acid, so with $\mathrm{NaClO_4}$ it has no buffering action. $\mathrm{CH_3COOH/CH_3COONa}$ , $\mathrm{H_2CO_3/Na_2CO_3}$ and $\mathrm{H_3PO_4/Na_3PO_4}$ are weak acid with a salt of that acid, so they are accepted as acidic buffers.

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2015 · NEET-UG (India) · 4 marks

Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified $\mathrm{KMnO_4}$ for complete oxidation? (a) $\mathrm{Fe(NO_2)_2}$ (b) $\mathrm{FeSO_4}$ (c) $\mathrm{FeSO_3}$ (d) $\mathrm{FeC_2O_4}$

Answer

(b) $\mathrm{FeSO_4}$

Count electrons lost per mole: $\mathrm{Fe(NO_2)_2}$ : $\mathrm{Fe^{2+}}\rightarrow\mathrm{Fe^{3+}}$ (1) plus $2\,\mathrm{NO_2^-}\rightarrow2\,\mathrm{NO_3^-}$ (2) $=3$ . $\mathrm{FeSO_3}$ : $1+2$ ( $\mathrm{SO_3^{2-}}\rightarrow\mathrm{SO_4^{2-}}$ ) $=3$ ; $\mathrm{FeC_2O_4}$ : $1+2$ ( $\mathrm{C_2O_4^{2-}}\rightarrow2\mathrm{CO_2}$ ) $=3$ . $\mathrm{FeSO_4}$ loses only 1 electron (sulphate is not oxidised), so it needs the least $\mathrm{KMnO_4}$ .

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2015 · NEET-UG (India) · 4 marks

The heat of combustion of carbon to $\mathrm{CO_2}$ is $-393.5$ kJ/mol. The heat released upon formation of 35.2 g of $\mathrm{CO_2}$ from carbon and oxygen gas is: (a) $-3.15$ kJ (b) $-315$ kJ (c) $+315$ kJ (d) $-630$ kJ

Answer

(b) $-315$ kJ

Formation of $\mathrm{CO_2}$ from C and $\mathrm{O_2}$ is the combustion of carbon, $\Delta H=-393.5$ kJ/mol. Moles of $\mathrm{CO_2}=35.2/44=0.8$ mol. Heat released $=0.8\times(-393.5)=-314.8\approx-315$ kJ. The sign is negative because heat is released, so (c) is wrong.

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2015 · NEET-UG (India) · 4 marks

The rate constant of the reaction $\mathrm{A\rightarrow B}$ is $0.6\times10^{-3}$ mole per second. If the concentration of A is 5 M, then concentration of B after 20 minutes is: (a) 0.72 M (b) 1.08 M (c) 3.60 M (d) 0.36 M

Answer

(a) 0.72 M

The unit of $k$ (mol per second) shows a zero-order reaction, for which $[B]=kt$ . $t=20\times60=1200$ s, so $[B]=0.6\times10^{-3}\times1200=0.72$ M, which is less than the 5 M available, so the reaction has not finished.

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2015 · NEET-UG (India) · 4 marks

If the equilibrium constant for $\mathrm{N_2(g)+O_2(g)\rightleftharpoons 2NO(g)}$ is $K$ , the equilibrium constant for $\tfrac12\mathrm{N_2(g)}+\tfrac12\mathrm{O_2(g)}\rightleftharpoons \mathrm{NO(g)}$ will be: (a) $K^2$ (b) $K^{1/2}$ (c) $\tfrac12 K$ (d) $K$

Answer

(b) $K^{1/2}$

For the reverse or scaled reaction the constant is raised to the same factor. Halving every coefficient of the reaction gives $K'=K^{1/2}$ . Check: $K=\dfrac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]}$ and $K'=\dfrac{[\mathrm{NO}]}{[\mathrm{N_2}]^{1/2}[\mathrm{O_2}]^{1/2}}=\sqrt K$ .

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2015 · NEET-UG (India) · 4 marks

Aqueous solution of which of the following compounds is the best conductor of electric current? (a) Fructose, $\mathrm{C_6H_{12}O_6}$ (b) Acetic acid, $\mathrm{C_2H_4O_2}$ (c) Hydrochloric acid, HCl (d) Ammonia, $\mathrm{NH_3}$

Answer

(c) Hydrochloric acid, HCl

HCl is a strong electrolyte that ionises completely in water, giving the highest concentration of ions and so the best conductivity. Acetic acid and ammonia are weak electrolytes (partial ionisation), and fructose is a non-electrolyte.

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2015 · NEET-UG (India) · 4 marks

What is the mole fraction of the solute in a 1.00 m aqueous solution? (a) 0.0177 (b) 0.177 (c) 1.770 (d) 0.0354

Answer

(a) 0.0177

1.00 m means 1 mol solute in 1000 g water, i.e. $1000/18=55.5$ mol water. $x=\dfrac{1}{1+55.5}=\dfrac{1}{56.5}=0.0177$ .

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2014 · NEET-UG (India) · 4 marks

Which of the following salts will give highest pH in water? (a) NaCl (b) $\mathrm{Na_2CO_3}$ (c) $\mathrm{CuSO_4}$ (d) KCl

Answer

(b) $\mathrm{Na_2CO_3}$

$\mathrm{Na_2CO_3}$ is the salt of a strong base and weak acid; $\mathrm{CO_3^{2-}}$ hydrolyses to give $\mathrm{OH^-}$ , so the solution is basic with the highest pH. NaCl and KCl are neutral (pH about 7) and $\mathrm{CuSO_4}$ is acidic.

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2014 · NEET-UG (India) · 4 marks

For the reaction: $\mathrm{X_2O_4(l)\rightarrow 2XO_2(g)}$ , $\Delta U=2.1$ kcal, $\Delta S=20\ \mathrm{cal\ K^{-1}}$ at 300 K. Hence $\Delta G$ is: (a) $-2.7$ kcal (b) 9.3 kcal (c) $-9.3$ kcal (d) 2.7 kcal

Answer

(a) $-2.7$ kcal

$\Delta n_g=2$ , so $\Delta H=\Delta U+\Delta n_gRT=2.1+2\times(2\times10^{-3})\times300=3.3$ kcal. $\Delta G=\Delta H-T\Delta S=3.3-300\times0.020=3.3-6.0=-2.7$ kcal.

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2014 · NEET-UG (India) · 4 marks

Using the Gibbs energy change, $\Delta G^\circ=+63.3$ kJ, for the following reaction, $\mathrm{Ag_2CO_3(s)\rightleftharpoons 2Ag^+(aq)+CO_3^{2-}(aq)}$ , the $K_{sp}$ of $\mathrm{Ag_2CO_3(s)}$ in water at 25 °C is: ( $R=8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$ ) (a) $8.0\times10^{-12}$ (b) $2.9\times10^{-3}$ (c) $7.9\times10^{-2}$ (d) $3.2\times10^{-26}$

Answer

(a) $8.0\times10^{-12}$

$\Delta G^\circ=-RT\ln K_{sp}$ , so $\ln K_{sp}=-\dfrac{63300}{8.314\times298}=-25.55$ . $K_{sp}=e^{-25.55}\approx8.0\times10^{-12}$ . A positive $\Delta G^\circ$ means $K<1$ , which rules out (b) and (c).

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2014 · NEET-UG (India) · 4 marks

For a given exothermic reaction, $K_p$ and $K_p'$ are the equilibrium constants at temperatures $T_1$ and $T_2$ , respectively. Assuming that heat of reaction is constant in temperature range between $T_1$ and $T_2$ , it is readily observed that: (a) $K_p<K_p'$ (b) $K_p=K_p'$ (c) $K_p=\dfrac{1}{K_p'}$ (d) $K_p>K_p'$

Answer

(d) $K_p>K_p'$

For an exothermic reaction the van't Hoff equation gives $\ln\dfrac{K_p'}{K_p}=\dfrac{\Delta H}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)$ , which is negative for $T_2>T_1$ (the usual reading). So the constant falls on heating: $K_p>K_p'$ . This follows Le Chatelier's principle, since a higher temperature shifts an exothermic equilibrium backwards.

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2014 · NEET-UG (India) · 4 marks

The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of $\mathrm{O_2}$ at STP will be: (a) 10.8 g (b) 54.0 g (c) 108.0 g (d) 5.4 g

Answer

(c) 108.0 g

5600 mL $\mathrm{O_2}$ at STP $=0.25$ mol, and $\mathrm{O_2}$ needs 4 $e^-$ per mole, so charge $=1$ mol of electrons (1 F). By Faraday's law, 1 mol $e^-$ deposits 1 mol of Ag ( $\mathrm{Ag^++e^-\rightarrow Ag}$ ) $=108$ g.

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2014 · NEET-UG (India) · 4 marks

For the reversible reaction: $\mathrm{N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+heat}$ . The equilibrium shifts in forward direction: (a) by decreasing the pressure (b) by decreasing the concentrations of $\mathrm{N_2(g)}$ and $\mathrm{H_2(g)}$ (c) by increasing pressure and decreasing temperature (d) by increasing the concentration of $\mathrm{NH_3(g)}$

Answer

(c) by increasing pressure and decreasing temperature

The forward reaction reduces gas moles (4 to 2) and is exothermic. By Le Chatelier's principle, higher pressure and lower temperature both favour the forward reaction. Lowering the pressure or the concentration of reactants, or raising $\mathrm{NH_3}$ , shifts it backwards.

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2014 · NEET-UG (India) · 4 marks

Equal masses of $\mathrm{H_2}$ , $\mathrm{O_2}$ and methane have been taken in a container of volume V at temperature 27 °C in identical conditions. The ratio of the volumes of gases $\mathrm{H_2:O_2:}$ methane would be: (a) 16 : 8 : 1 (b) 16 : 1 : 2 (c) 8 : 1 : 2 (d) 8 : 16 : 1

Answer

(b) 16 : 1 : 2

For equal mass $w$ , moles and volume (at the same T and P) are proportional to $1/M$ . $V \propto \frac12 : \frac1{32} : \frac1{16} = 16:1:2$ for $\mathrm{H_2:O_2:CH_4}$ .

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2014 · NEET-UG (India) · 4 marks

When 0.1 mol $\mathrm{MnO_4^{2-}}$ is oxidised the quantity of electricity required to completely oxidise $\mathrm{MnO_4^{2-}}$ to $\mathrm{MnO_4^-}$ is: (a) $2\times96500$ C (b) 9650 C (c) 96.50 C (d) 96500 C

Answer

(b) 9650 C

The change is $\mathrm{MnO_4^{2-}\rightarrow MnO_4^-}+e^-$ , so one electron per ion. $Q = nF = 0.1\times96500 = 9650$ C.

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2014 · NEET-UG (India) · 4 marks

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? (a) $\mathrm{C_6H_{12}O_6}$ (b) $\mathrm{Al_2(SO_4)_3}$ (c) $\mathrm{K_2SO_4}$ (d) KCl

Answer

(b) $\mathrm{Al_2(SO_4)_3}$

$\Delta T_f = i K_f m$ is a colligative property, so it depends on the van't Hoff factor $i$ at equal molality. $\mathrm{Al_2(SO_4)_3}$ gives $2\mathrm{Al^{3+}}+3\mathrm{SO_4^{2-}}$ , so $i=5$ , against 3 for $\mathrm{K_2SO_4}$ , 2 for KCl and 1 for glucose.

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2013 · NEET-UG (India) · 4 marks

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20 °C to 35 °C? ( $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$ ) (a) $342\ \mathrm{kJ\ mol^{-1}}$ (b) $269\ \mathrm{kJ\ mol^{-1}}$ (c) $34.7\ \mathrm{kJ\ mol^{-1}}$ (d) $15.1\ \mathrm{kJ\ mol^{-1}}$

Answer

(c) $34.7\ \mathrm{kJ\ mol^{-1}}$

Rate doubles, so $k_2/k_1=2$ , with $T_1=293$ K and $T_2=308$ K. $E_a=\dfrac{2.303R\,T_1T_2}{T_2-T_1}\log\dfrac{k_2}{k_1}=\dfrac{8.314\times0.693\times293\times308}{15}\approx3.47\times10^4$ J $=34.7$ kJ/mol.

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2013 · NEET-UG (India) · 4 marks

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be (a) 0.059 V (b) 0.59 V (c) 0.118 V (d) 1.18 V

Answer

(b) 0.59 V

For $\mathrm{H_2\rightarrow2H^++2e^-}$ at 1 atm, $E_{ox}=-0.059\log[\mathrm{H^+}]$ (since $E^\circ=0$ and $n=2$ gives $0.0591\times\frac{2}{2}$ ). $E_{ox}=0.059\times\mathrm{pH}=0.059\times10=0.59$ V.

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2013 · NEET-UG (India) · 4 marks

A reaction having equal energies of activation for forward and reverse reactions has (a) $\Delta S=0$ (b) $\Delta G=0$ (c) $\Delta H=0$ (d) $\Delta H=\Delta G=\Delta S=0$

Answer

(c) $\Delta H=0$

$\Delta H = E_{a(f)}-E_{a(r)}$ . If the two activation energies are equal then $\Delta H=0$ . This says nothing about $\Delta S$ or $\Delta G$ , which depend on other factors.

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2013 · NEET-UG (India) · 4 marks

At 25 °C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9.54\ \mathrm{ohm^{-1}\ cm^2\ mol^{-1}}$ and at infinite dilution its molar conductance is $238\ \mathrm{ohm^{-1}\ cm^2\ mol^{-1}}$ . The degree of ionisation of ammonium hydroxide at the same concentration and temperature is (a) 2.080% (b) 20.800% (c) 4.008% (d) 40.800%

Answer

(c) 4.008%

$\alpha=\Lambda_m/\Lambda_m^\circ=\dfrac{9.54}{238}=0.04008$ . So the degree of ionisation is 4.008%.

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2013 · NEET-UG (India) · 4 marks

A button cell used in watches functions as following: $\mathrm{Zn(s)+Ag_2O(s)+H_2O(l)\rightleftharpoons 2Ag(s)+Zn^{2+}(aq)+2OH^-(aq)}$ If half cell potentials are $\mathrm{Zn^{2+}(aq)+2e^-\rightarrow Zn(s)}$ ; $E^\circ=-0.76$ V $\mathrm{Ag_2O(s)+H_2O(l)+2e^-\rightarrow 2Ag(s)+2OH^-(aq)}$ ; $E^\circ=0.34$ V The cell potential will be (a) 1.10 V (b) 0.42 V (c) 0.84 V (d) 1.34 V

Answer

(a) 1.10 V

$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ . Ag $_2$ O is reduced (cathode, 0.34 V) and Zn is oxidised (anode, $-0.76$ V). $E^\circ_{cell}=0.34-(-0.76)=1.10$ V.

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2013 · NEET-UG (India) · 4 marks

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M $\mathrm{HNO_3}$ ? The concentrated acid is 70% $\mathrm{HNO_3}$ . (a) 45.0 g conc. $\mathrm{HNO_3}$ (b) 90.0 g conc. $\mathrm{HNO_3}$ (c) 70.0 g conc. $\mathrm{HNO_3}$ (d) 54.0 g conc. $\mathrm{HNO_3}$

Answer

(a) 45.0 g conc. $\mathrm{HNO_3}$

Moles needed $=2.0\times0.250=0.5$ mol; mass of pure $\mathrm{HNO_3}=0.5\times63=31.5$ g. The acid is 70% pure, so mass of solution $=31.5/0.70=45.0$ g.

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2013 · NEET-UG (India) · 4 marks

Maximum deviation from ideal gas is expected from (a) $\mathrm{H_2(g)}$ (b) $\mathrm{N_2(g)}$ (c) $\mathrm{CH_4(g)}$ (d) $\mathrm{NH_3(g)}$

Answer

(d) $\mathrm{NH_3(g)}$

Deviation from ideality is largest for the gas with the strongest intermolecular forces and easiest liquefaction. $\mathrm{NH_3}$ is polar and hydrogen bonded, with the highest critical temperature and van der Waals constant $a$ of these gases. $\mathrm{H_2}$ and $\mathrm{N_2}$ are nearly ideal.

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2013 · NEET-UG (India) · 4 marks

Dipole-induced dipole interactions are present in which of the following pairs? (a) $\mathrm{H_2O}$ and alcohol (b) $\mathrm{Cl_2}$ and $\mathrm{CCl_4}$ (c) HCl and He atoms (d) $\mathrm{SiF_4}$ and He atoms

Answer

(c) HCl and He atoms

A dipole-induced dipole interaction needs one polar molecule and one non-polar species that can be polarised. HCl is polar and He is a non-polar atom. In (a) both are polar, in (b) both are non-polar, and in (d) $\mathrm{SiF_4}$ is non-polar too (it gives only dispersion forces with He).

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2013 · NEET-UG (India) · 4 marks

$6.02\times10^{20}$ molecules of urea are present in 100 mL of its solution. The concentration of solution is (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M

Answer

(b) 0.01 M

Moles of urea $=\dfrac{6.02\times10^{20}}{6.02\times10^{23}}=10^{-3}$ mol. Molarity $=\dfrac{10^{-3}\ \text{mol}}{0.1\ \text{L}}=0.01\ \text{M}$ . The common slip is forgetting to convert 100 mL to 0.1 L, which gives 0.001 M (c).

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2017 · NEET-UG (India) · 4 marks

Concentration of the $\mathrm{Ag^+}$ ions in a saturated solution of $\mathrm{Ag_2C_2O_4}$ is $2.2\times10^{-4}\ \mathrm{mol\,L^{-1}}$ . Solubility product of $\mathrm{Ag_2C_2O_4}$ is
(a) $2.42\times10^{-8}$ (b) $2.66\times10^{-12}$ (c) $4.5\times10^{-11}$ (d) $5.3\times10^{-12}$

Answer

(d) $5.3\times10^{-12}$

$\mathrm{Ag_2C_2O_4 \rightleftharpoons 2Ag^+ + C_2O_4^{2-}}$ . Given $[\mathrm{Ag^+}] = 2.2\times10^{-4}$ M, so $[\mathrm{C_2O_4^{2-}}] = 1.1\times10^{-4}$ M.

$K_{sp} = [\mathrm{Ag^+}]^2[\mathrm{C_2O_4^{2-}}] = (2.2\times10^{-4})^2(1.1\times10^{-4}) = 5.3\times10^{-12}$ .

Using $[\mathrm{Ag^+}]$ for the oxalate concentration, or not squaring $[\mathrm{Ag^+}]$ , gives the wrong options.

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2017 · NEET-UG (India) · 4 marks

If molality of the dilute solution is doubled, the value of molal depression constant ( $K_f$ ) will be
(a) Doubled (b) Halved (c) Tripled (d) Unchanged

Answer

(d) Unchanged

The molal depression constant $K_f = \dfrac{RT_f^2 M_1}{1000\,\Delta H_{fus}}$ depends only on the solvent (its freezing point, molar mass and enthalpy of fusion). It does not depend on the concentration of the solute, so doubling the molality doubles $\Delta T_f$ but leaves $K_f$ unchanged.

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2017 · NEET-UG (India) · 4 marks

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy $\Delta U$ of the gas in joules will be
(a) 1136.25 J (b) $-500$ J (c) $-505$ J (d) $+505$ J

Answer

(c) $-505$ J

The container is insulated, so $q = 0$ and $\Delta U = w = -P_{ext}\Delta V$ .

$\Delta V = 4.50 - 2.50 = 2.00$ L, so $w = -2.5 \times 2.00 = -5.0$ L atm.

Using 1 L atm = 101.3 J, $\Delta U \approx -506.5$ J, which is closest to $-505$ J. The sign is negative because the gas does work on the surroundings and loses internal energy, so the positive options are wrong.

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2017 · NEET-UG (India) · 4 marks

In the electrochemical cell: $\mathrm{Zn|ZnSO_4(0.01\ M)||CuSO_4(1.0\ M)|Cu}$ , the emf of this Daniel cell is $E_1$ . When the concentration of $\mathrm{ZnSO_4}$ is changed to 1.0 M and that of $\mathrm{CuSO_4}$ changed to 0.01 M, the emf changes to $E_2$ . From the following, which one is the relationship between $E_1$ and $E_2$ ? (Given, $\dfrac{RT}{F} = 0.059$ )
(a) $E_1 = E_2$ (b) $E_1 < E_2$ (c) $E_1 > E_2$ (d) $E_2 = 0 \neq E_1$

Answer

(c) $E_1 > E_2$

For $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$ , $n=2$ and $E = E^\circ - \dfrac{0.059}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}$ .

$E_1 = E^\circ - 0.0295\log(0.01/1) = E^\circ + 0.059$ .

$E_2 = E^\circ - 0.0295\log(1/0.01) = E^\circ - 0.059$ .

So $E_1 > E_2$ .

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2017 · NEET-UG (India) · 4 marks

A 20 litre container at 400 K contains $\mathrm{CO_2(g)}$ at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of $\mathrm{CO_2}$ attains its maximum value, will be (Given that: $\mathrm{SrCO_3(s)} \rightleftharpoons \mathrm{SrO(s)} + \mathrm{CO_2(g)}$ , $K_p = 1.6$ atm)
(a) 5 litre (b) 10 litre (c) 4 litre (d) 2 litre

Answer

(a) 5 litre

The equilibrium $\mathrm{SrCO_3 \rightleftharpoons SrO + CO_2}$ has $K_p = p_{\mathrm{CO_2}} = 1.6$ atm, which is the maximum pressure of $\mathrm{CO_2}$ . Until this is reached no reaction occurs, so Boyle's law applies: $0.4 \times 20 = 1.6 \times V$ , giving $V = 5$ L. Compressing beyond this makes $\mathrm{CO_2}$ recombine with SrO rather than raising the pressure.

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2017 · NEET-UG (India) · 4 marks

Which of the following is dependent on temperature?
(a) Molality (b) Molarity (c) Mole fraction (d) Weight percentage

Answer

(b) Molarity

Molarity is moles per litre of solution, and the solution volume changes with temperature, so molarity does too. Molality, mole fraction and weight percentage involve only masses or moles, which do not change with temperature.

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2017 · NEET-UG (India) · 4 marks

For a given reaction, $\Delta H = 35.5\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S = 83.6\ \mathrm{J\,K^{-1}mol^{-1}}$ . The reaction is spontaneous at: (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)
(a) $T < 425$ K (b) $T > 425$ K (c) All temperatures (d) $T > 298$ K

Answer

(b) $T > 425$ K

A reaction is spontaneous when $\Delta G = \Delta H - T\Delta S < 0$ . With $\Delta H$ and $\Delta S$ both positive, this needs $T > \Delta H/\Delta S$ .

$T > \dfrac{35.5\times10^3}{83.6} = 424.6 \approx 425$ K.

At lower temperatures the unfavourable enthalpy term dominates, so (a) is wrong.

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2017 · NEET-UG (India) · 4 marks

A first order reaction has a specific reaction rate of $10^{-2}\ \mathrm{s^{-1}}$ . How much time will it take for 20 g of the reactant to reduce to 5 g?
(a) 238.6 second (b) 138.6 second (c) 346.5 second (d) 693.0 second

Answer

(b) 138.6 second

For first order, $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$ .

$t = \dfrac{2.303}{10^{-2}}\log\dfrac{20}{5} = 230.3 \times 0.602 = 138.6$ s.

This is two half-lives ( $20 \rightarrow 10 \rightarrow 5$ g), and $t_{1/2} = 0.693/k = 69.3$ s, so $2\times69.3 = 138.6$ s.

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2017 · NEET-UG (India) · 4 marks

The equilibrium constants of the following are:
$\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$ , $K_1$ $\mathrm{N_2} + \mathrm{O_2} \rightleftharpoons 2\mathrm{NO}$ , $K_2$ $\mathrm{H_2} + \tfrac{1}{2}\mathrm{O_2} \rightarrow \mathrm{H_2O}$ , $K_3$
The equilibrium constant ( $K$ ) of the reaction $2\mathrm{NH_3} + \tfrac{5}{2}\mathrm{O_2} \rightleftharpoons 2\mathrm{NO} + 3\mathrm{H_2O}$ will be
(a) $K_1K_3^3/K_2$ (b) $K_2K_3^3/K_1$ (c) $K_2K_3/K_1$ (d) $K_2^3K_3/K_1$

Answer

(b) $K_2K_3^3/K_1$

Build the target reaction $2\mathrm{NH_3} + \tfrac52\mathrm{O_2} \rightleftharpoons 2\mathrm{NO} + 3\mathrm{H_2O}$ from the given ones:

reverse reaction 1: $K = 1/K_1$
add reaction 2: $K_2$
add 3 times reaction 3: $K_3^3$

Multiplying: $K = \dfrac{K_2K_3^3}{K_1}$ .

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2017 · NEET-UG (India) · 4 marks

Mechanism of a hypothetical reaction $\mathrm{X_2} + \mathrm{Y_2} \rightarrow 2\mathrm{XY}$ is given below:
(i) $\mathrm{X_2} \rightarrow \mathrm{X} + \mathrm{X}$ (fast) (ii) $\mathrm{X} + \mathrm{Y_2} \rightleftharpoons \mathrm{XY} + \mathrm{Y}$ (slow) (iii) $\mathrm{X} + \mathrm{Y} \rightarrow \mathrm{XY}$ (fast)
The overall order of the reaction will be
(a) 1 (b) 2 (c) 0 (d) 1.5

Answer

(d) 1.5

The slowest step decides the rate: rate = $k[\mathrm{X}][\mathrm{Y_2}]$ . X is an intermediate, so use the fast equilibrium in step (i): $[\mathrm{X}] = K^{1/2}[\mathrm{X_2}]^{1/2}$ .

Rate = $k'[\mathrm{X_2}]^{1/2}[\mathrm{Y_2}]$ , so the overall order is $\tfrac12 + 1 = 1.5$ .

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2018 · NEET-UG (India) · 4 marks

Given van der Waals constant $a$ for $\mathrm{NH_3}$ , $\mathrm{H_2}$ , $\mathrm{O_2}$ and $\mathrm{CO_2}$ are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
(a) $\mathrm{NH_3}$ (b) $\mathrm{O_2}$ (c) $\mathrm{H_2}$ (d) $\mathrm{CO_2}$

Answer

(a) $\mathrm{NH_3}$

The van der Waals constant $a$ measures the strength of intermolecular attractions. A larger $a$ means stronger attraction, so the gas is liquefied more easily. $\mathrm{NH_3}$ has the largest $a$ (4.17), so it is most easily liquefied. $\mathrm{H_2}$ (0.244) is the hardest.

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2018 · NEET-UG (India) · 4 marks

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL $\dfrac{M}{10}$ HCl + 40 mL $\dfrac{M}{10}$ NaOH b. 55 mL $\dfrac{M}{10}$ HCl + 45 mL $\dfrac{M}{10}$ NaOH c. 75 mL $\dfrac{M}{5}$ HCl + 25 mL $\dfrac{M}{5}$ NaOH d. 100 mL $\dfrac{M}{10}$ HCl + 100 mL $\dfrac{M}{10}$ NaOH
pH of which one of them will be equal to 1?
(a) b (b) d (c) a (d) c

Answer

(d) c

pH = 1 needs $[\mathrm{H^+}] = 0.1$ M after mixing (excess acid).

a: $6 - 4 = 2$ mmol in 100 mL = 0.02 M, so pH is about 1.7
b: $5.5 - 4.5 = 1$ mmol in 100 mL = 0.01 M, so pH = 2
c: $75\times0.2 - 25\times0.2 = 10$ mmol in 100 mL = 0.1 M, so pH = 1
d: equal amounts, neutral, pH = 7

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2018 · NEET-UG (India) · 4 marks

The solubility of $\mathrm{BaSO_4}$ in water is $2.42\times10^{-3}\ \mathrm{g\,L^{-1}}$ at 298 K. The value of its solubility product ( $K_{sp}$ ) will be (Given molar mass of $\mathrm{BaSO_4}$ = 233 g $\mathrm{mol^{-1}}$ )
(a) $1.08\times10^{-10}\ \mathrm{mol^2L^{-2}}$ (b) $1.08\times10^{-14}\ \mathrm{mol^2L^{-2}}$ (c) $1.08\times10^{-12}\ \mathrm{mol^2L^{-2}}$ (d) $1.08\times10^{-8}\ \mathrm{mol^2L^{-2}}$

Answer

(a) $1.08\times10^{-10}\ \mathrm{mol^2L^{-2}}$

Solubility $s = \dfrac{2.42\times10^{-3}}{233} = 1.04\times10^{-5}$ mol/L.

$\mathrm{BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}}$ , so $K_{sp} = s^2 = (1.04\times10^{-5})^2 = 1.08\times10^{-10}$ .

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2018 · NEET-UG (India) · 4 marks

The correct difference between first and second order reactions is that
(a) The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations (b) A first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed (c) The half-life of a first-order reaction does not depend on $[A]_0$ ; the half-life of a second-order reaction does depend on $[A]_0$ (d) The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

Answer

(c) The half-life of a first-order reaction does not depend on $[A]_0$ ; the half-life of a second-order reaction does depend on $[A]_0$

For first order, $t_{1/2}=0.693/k$ , independent of initial concentration. For second order, $t_{1/2}=1/(k[A]_0)$ , which depends on $[A]_0$ . Options (a) and (d) are wrong because both orders depend on concentration, and (b) is wrong because catalysis is possible for either.

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2018 · NEET-UG (India) · 4 marks

The bond dissociation energies of $\mathrm{X_2}$ , $\mathrm{Y_2}$ and XY are in the ratio of 1 : 0.5 : 1. $\Delta H$ for the formation of XY is $-200\ \mathrm{kJ\,mol^{-1}}$ . The bond dissociation energy of $\mathrm{X_2}$ will be
(a) $200\ \mathrm{kJ\,mol^{-1}}$ (b) $800\ \mathrm{kJ\,mol^{-1}}$ (c) $100\ \mathrm{kJ\,mol^{-1}}$ (d) $400\ \mathrm{kJ\,mol^{-1}}$

Answer

(b) $800\ \mathrm{kJ\,mol^{-1}}$

Let the bond energies of $\mathrm{X_2}$ , $\mathrm{Y_2}$ and XY be $a$ , $0.5a$ and $a$ . For $\tfrac12\mathrm{X_2}+\tfrac12\mathrm{Y_2}\rightarrow \mathrm{XY}$ : $\Delta H=\tfrac12 a+\tfrac12(0.5a)-a=-0.25a$ . Setting $-0.25a=-200$ gives $a=800\ \mathrm{kJ\,mol^{-1}}$ .

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2018 · NEET-UG (India) · 4 marks

When initial concentration of the reactant is doubled, the half-life period of a zero order reaction
(a) Is halved (b) Is tripled (c) Is doubled (d) Remains unchanged

Answer

(c) Is doubled

For a zero order reaction, $t_{1/2}=\dfrac{[A]_0}{2k}$ , so the half-life is directly proportional to the initial concentration. Doubling $[A]_0$ doubles $t_{1/2}$ . Only first order reactions have a half-life that is unchanged.

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2018 · NEET-UG (India) · 4 marks

For the redox reaction $\mathrm{MnO_4^-} + \mathrm{C_2O_4^{2-}} + \mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + \mathrm{CO_2} + \mathrm{H_2O}$ , the correct coefficients of the reactants for the balanced equation are, in the order $\mathrm{MnO_4^-}$ , $\mathrm{C_2O_4^{2-}}$ , $\mathrm{H^+}$ :
(a) 16, 5, 2 (b) 2, 16, 5 (c) 2, 5, 16 (d) 5, 16, 2

Answer

(c) 2, 5, 16

Half reactions: $\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}$ and $\mathrm{C_2O_4^{2-}\rightarrow 2CO_2+2e^-}$ . Multiply the first by 2 and the second by 5 to equalise 10 electrons. This gives $2\mathrm{MnO_4^-}+5\mathrm{C_2O_4^{2-}}+16\mathrm{H^+}\rightarrow 2\mathrm{Mn^{2+}}+10\mathrm{CO_2}+8\mathrm{H_2O}$ .

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2018 · NEET-UG (India) · 4 marks

Which one of the following conditions will favour maximum formation of the product in the reaction $\mathrm{A_2(g)} + \mathrm{B_2(g)} \rightleftharpoons \mathrm{X_2(g)}$ , $\Delta_r H = -X\ \mathrm{kJ}$ ?
(a) Low temperature and high pressure (b) High temperature and high pressure (c) Low temperature and low pressure (d) High temperature and low pressure

Answer

(a) Low temperature and high pressure

The forward reaction is exothermic ( $\Delta_rH<0$ ), so by Le Chatelier's principle low temperature favours the product. It also reduces the number of gas moles (2 mol to 1 mol), so high pressure shifts the equilibrium to the product side. High temperature would favour the reverse reaction.

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2018 · NEET-UG (India) · 4 marks

The correction factor 'a' to the ideal gas equation corresponds to
(a) Density of the gas molecules (b) Electric field present between the gas molecules (c) Volume of the gas molecules (d) Forces of attraction between the gas molecules

Answer

(d) Forces of attraction between the gas molecules

In the van der Waals equation $\left(P+\dfrac{an^2}{V^2}\right)(V-nb)=nRT$ , the term $a$ corrects the pressure for intermolecular attractive forces. The constant $b$ , not $a$ , corrects for the finite volume of molecules, which is why (c) is wrong.

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2019 · NEET-UG (India) · 4 marks

For a cell involving one electron $E^\circ_{cell} = 0.59\ \mathrm{V}$ at 298 K, the equilibrium constant for the cell reaction is: [Given that $\dfrac{2.303RT}{F} = 0.059\ \mathrm{V}$ at $T = 298\ \mathrm{K}$ ]
(a) $1.0\times10^{5}$ (b) $1.0\times10^{10}$ (c) $1.0\times10^{30}$ (d) $1.0\times10^{2}$

Answer

(b) $1.0\times10^{10}$

At 298 K, $E^\circ_{cell}=\dfrac{0.059}{n}\log K$ . With $n=1$ : $\log K=\dfrac{0.59}{0.059}=10$ , so $K=10^{10}$ .

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2019 · NEET-UG (India) · 4 marks

A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor ( $Z$ ) is:
(a) $Z>1$ and repulsive forces are dominant (b) $Z<1$ and attractive forces are dominant (c) $Z<1$ and repulsive forces are dominant (d) $Z>1$ and attractive forces are dominant

Answer

(b) $Z<1$ and attractive forces are dominant

$Z=\dfrac{V_{real}}{V_{ideal}}=0.8$ since the molar volume is 20 percent smaller. $Z<1$ means the gas is more compressible than ideal, which happens when intermolecular attractions dominate. $Z>1$ would indicate dominant repulsion.

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2019 · NEET-UG (India) · 4 marks

Which of the following reactions are disproportionation reaction?
(a) $2\mathrm{Cu^+} \rightarrow \mathrm{Cu^{2+}} + \mathrm{Cu^0}$ (b) $3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$ (c) $2\mathrm{KMnO_4} \rightarrow \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}$ (d) $2\mathrm{MnO_4^-} + 3\mathrm{Mn^{2+}} + 2\mathrm{H_2O} \rightarrow 5\mathrm{MnO_2} + 4\mathrm{H^+}$
Select the correct option from the following
(a) (a), (b) and (c) (b) (a), (c) and (d) (c) (a) and (d) only (d) (a) and (b) only

Answer

(d) (a) and (b) only

A disproportionation reaction has the same species oxidised and reduced. In (a), $\mathrm{Cu^+}$ goes to $\mathrm{Cu^{2+}}$ and $\mathrm{Cu^0}$ . In (b), $\mathrm{Mn(VI)}$ goes to $\mathrm{Mn(VII)}$ and $\mathrm{Mn(IV)}$ . Reaction (c) is a decomposition with separate species changing, and (d) is comproportionation, where $\mathrm{Mn(VII)}$ and $\mathrm{Mn(II)}$ both form $\mathrm{Mn(IV)}$ .

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2019 · NEET-UG (India) · 4 marks

Which will make basic buffer?
(a) 100 mL of 0.1 M $\mathrm{CH_3COOH}$ + 100 mL of 0.1 M NaOH (b) 100 mL of 0.1 M HCl + 200 mL of 0.1 M $\mathrm{NH_4OH}$ (c) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH (d) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M $\mathrm{CH_3COOH}$

Answer

(b) 100 mL of 0.1 M HCl + 200 mL of 0.1 M $\mathrm{NH_4OH}$

A basic buffer is a weak base plus its salt. Here 10 mmol HCl reacts with 20 mmol $\mathrm{NH_4OH}$ to give 10 mmol $\mathrm{NH_4Cl}$ with 10 mmol $\mathrm{NH_4OH}$ left over, which is a buffer. Options (a) and (c) give a salt only (or strong acid and base neutralised), and (d) leaves excess strong base with the weak acid fully neutralised, not a basic buffer.

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2019 · NEET-UG (India) · 4 marks

Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (Given that 1 L bar = 100 J)
(a) 5 kJ (b) 25 J (c) 30 J (d) $-30$ J

Answer

(d) $-30$ J

For expansion against constant external pressure, $w = -p_{ext}\Delta V = -2\ \text{bar}\times(0.25-0.1)\ \text{L} = -0.3\ \text{L bar} = -30$ J. The sign is negative because the gas does work on the surroundings. The temperature value is irrelevant here, since the formula uses only $p_{ext}$ and $\Delta V$ .

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2019 · NEET-UG (India) · 4 marks

The mixture that forms maximum boiling azeotrope is:
(a) Ethanol + Water (b) Acetone + Carbon disulphide (c) Heptane + Octane (d) Water + Nitric acid

Answer

(d) Water + Nitric acid

Mixtures showing large negative deviation from Raoult's law form maximum boiling azeotropes, and $\mathrm{H_2O}$ + $\mathrm{HNO_3}$ is the standard example (68% $\mathrm{HNO_3}$ ). Ethanol + water, and acetone + $\mathrm{CS_2}$ , show positive deviation and form minimum boiling azeotropes. Heptane + octane is almost ideal.

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2019 · NEET-UG (India) · 4 marks

For an ideal solution, the correct option is:
(a) $\Delta_{mix}V \neq 0$ at constant T and P (b) $\Delta_{mix}H = 0$ at constant T and P (c) $\Delta_{mix}G = 0$ at constant T and P (d) $\Delta_{mix}S = 0$ at constant T and P

Answer

(b) $\Delta_{mix}H = 0$ at constant T and P

In an ideal solution the A-B interactions equal the A-A and B-B interactions. So there is no heat change ( $\Delta_{mix}H=0$ ) and no volume change ( $\Delta_{mix}V=0$ ). Mixing is spontaneous with $\Delta_{mix}S>0$ , so $\Delta_{mix}G<0$ , not zero.

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2019 · NEET-UG (India) · 4 marks

If the rate constant for a first order reaction is $k$ , the time ( $t$ ) required for the completion of 99% of the reaction is given by:
(a) $t = 6.909/k$ (b) $t = 4.606/k$ (c) $t = 2.303/k$ (d) $t = 0.693/k$

Answer

(b) $t = 4.606/k$

For a first order reaction, $t=\dfrac{2.303}{k}\log\dfrac{a}{a-x}$ . For 99% completion, $a-x=0.01a$ , so $t=\dfrac{2.303}{k}\log 100=\dfrac{2.303\times2}{k}=\dfrac{4.606}{k}$ .

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2019 · NEET-UG (India) · 4 marks

Conjugate base for Brønsted acids $\mathrm{H_2O}$ and HF are:
(a) $\mathrm{H_3O^+}$ and $\mathrm{F^-}$ , respectively (b) $\mathrm{OH^-}$ and $\mathrm{F^-}$ , respectively (c) $\mathrm{H_3O^+}$ and $\mathrm{H_2F^+}$ , respectively (d) $\mathrm{OH^-}$ and $\mathrm{H_2F^+}$ , respectively

Answer

(b) $\mathrm{OH^-}$ and $\mathrm{F^-}$ , respectively

A conjugate base is the species left after the acid loses one $\mathrm{H^+}$ . $\mathrm{H_2O}-\mathrm{H^+}=\mathrm{OH^-}$ and $\mathrm{HF}-\mathrm{H^+}=\mathrm{F^-}$ . Options with $\mathrm{H_3O^+}$ or $\mathrm{H_2F^+}$ describe conjugate acids, which have gained a proton.

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2019 · NEET-UG (India) · 4 marks

For the chemical reaction $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$ , the correct option is:
(a) $-\dfrac{d[\mathrm{N_2}]}{dt} = 2\dfrac{d[\mathrm{NH_3}]}{dt}$ (b) $-\dfrac{d[\mathrm{N_2}]}{dt} = \dfrac{1}{2}\dfrac{d[\mathrm{NH_3}]}{dt}$ (c) $-3\dfrac{d[\mathrm{H_2}]}{dt} = 2\dfrac{d[\mathrm{NH_3}]}{dt}$ (d) $-\dfrac{1}{3}\dfrac{d[\mathrm{H_2}]}{dt} = -\dfrac{1}{2}\dfrac{d[\mathrm{NH_3}]}{dt}$

Answer

(b) $-\dfrac{d[\mathrm{N_2}]}{dt}=\dfrac{1}{2}\dfrac{d[\mathrm{NH_3}]}{dt}$

The rate of reaction is $r=-\dfrac{d[\mathrm{N_2}]}{dt}=-\dfrac13\dfrac{d[\mathrm{H_2}]}{dt}=+\dfrac12\dfrac{d[\mathrm{NH_3}]}{dt}$ . Reactants fall (negative sign), products rise, and each rate is divided by its stoichiometric coefficient. Option (d) is wrong because it gives $\mathrm{NH_3}$ a negative sign.

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2019 · NEET-UG (India) · 4 marks

In which case change in entropy is negative?
(a) Expansion of a gas at constant temperature (b) Sublimation of solid to gas (c) $2\mathrm{H(g)} \rightarrow \mathrm{H_2(g)}$ (d) Evaporation of water

Answer

(c) $2\mathrm{H(g)}\rightarrow\mathrm{H_2(g)}$

Two moles of gas become one mole, so disorder decreases and $\Delta S<0$ . Gas expansion, sublimation and evaporation all increase disorder, so $\Delta S>0$ for each.

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2019 · NEET-UG (India) · 4 marks

For the cell reaction $2\mathrm{Fe^{3+}(aq)} + 2\mathrm{I^-(aq)} \rightarrow 2\mathrm{Fe^{2+}(aq)} + \mathrm{I_2(aq)}$ , $E^\ominus_{cell} = 0.24\ \mathrm{V}$ at 298 K. The standard Gibbs energy ( $\Delta_r G^\ominus$ ) of the cell reaction is: [Given that Faraday constant $F = 96500\ \mathrm{C\,mol^{-1}}$ ]
(a) $-23.16\ \mathrm{kJ\,mol^{-1}}$ (b) $46.32\ \mathrm{kJ\,mol^{-1}}$ (c) $23.16\ \mathrm{kJ\,mol^{-1}}$ (d) $-46.32\ \mathrm{kJ\,mol^{-1}}$

Answer

(d) $-46.32\ \mathrm{kJ\,mol^{-1}}$

$\Delta_rG^\ominus=-nFE^\ominus_{cell}$ . The reaction as written transfers $n=2$ electrons, so $\Delta_rG^\ominus=-2\times96500\times0.24=-46320\ \mathrm{J}=-46.32$ kJ/mol. Option (a) comes from wrongly using $n=1$ .

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2019 · NEET-UG (India) · 4 marks

pH of a saturated solution of $\mathrm{Ca(OH)_2}$ is 9. The solubility product ( $K_{sp}$ ) of $\mathrm{Ca(OH)_2}$ is:
(a) $0.25\times10^{-10}$ (b) $0.125\times10^{-15}$ (c) $0.5\times10^{-10}$ (d) $0.5\times10^{-15}$

Answer

(d) $0.5\times10^{-15}$

pH = 9 gives pOH = 5, so $[\mathrm{OH^-}]=10^{-5}$ M and $[\mathrm{Ca^{2+}}]=\tfrac12\times10^{-5}=5\times10^{-6}$ M. $K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2=5\times10^{-6}\times10^{-10}=5\times10^{-16}=0.5\times10^{-15}$ .

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2020 · NEET-UG (India) · 4 marks

The correct option for free expansion of an ideal gas under adiabatic condition is:
(a) $q < 0$ , $\Delta T = 0$ and $w = 0$ (b) $q > 0$ , $\Delta T > 0$ and $w > 0$ (c) $q = 0$ , $\Delta T = 0$ and $w = 0$ (d) $q = 0$ , $\Delta T < 0$ and $w > 0$

Answer

(c) $q=0$ , $\Delta T=0$ and $w=0$

Free expansion goes against zero external pressure, so $w=0$ . Adiabatic means $q=0$ . Then $\Delta U=q+w=0$ , and for an ideal gas $U$ depends only on $T$ , so $\Delta T=0$ .

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2020 · NEET-UG (India) · 4 marks

A mixture of $\mathrm{N_2}$ and Ar gases in a cylinder contains 7 g of $\mathrm{N_2}$ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of $\mathrm{N_2}$ is:
(a) 15 bar (b) 18 bar (c) 9 bar (d) 12 bar

Answer

(a) 15 bar

$n(\mathrm{N_2})=7/28=0.25$ mol and $n(\mathrm{Ar})=8/40=0.2$ mol. The mole fraction of $\mathrm{N_2}$ is $0.25/0.45=5/9$ . So $p_{\mathrm{N_2}}=\tfrac59\times27=15$ bar (Dalton's law).

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2020 · NEET-UG (India) · 4 marks

The rate constant for a first order reaction is $4.606\times10^{-3}\ \mathrm{s^{-1}}$ . The time required to reduce 2.0 g of the reactant to 0.2 g is:
(a) 500 s (b) 1000 s (c) 100 s (d) 200 s

Answer

(a) 500 s

For a first order reaction, $t=\dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}=\dfrac{2.303}{4.606\times10^{-3}}\log\dfrac{2.0}{0.2}=\dfrac{2.303}{4.606\times10^{-3}}\times1=500$ s.

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2020 · NEET-UG (India) · 4 marks

Hydrolysis of sucrose is given by the following reaction: Sucrose + $\mathrm{H_2O}$ $\rightleftharpoons$ Glucose + Fructose. If the equilibrium constant ( $K_c$ ) is $2\times10^{13}$ at 300 K, the value of $\Delta_r G^\ominus$ at the same temperature will be:
(a) $8.314\ \mathrm{J\,mol^{-1}K^{-1}}\times300\ \mathrm{K}\times\ln(3\times10^{13})$ (b) $-8.314\ \mathrm{J\,mol^{-1}K^{-1}}\times300\ \mathrm{K}\times\ln(4\times10^{13})$ (c) $-8.314\ \mathrm{J\,mol^{-1}K^{-1}}\times300\ \mathrm{K}\times\ln(2\times10^{13})$ (d) $8.314\ \mathrm{J\,mol^{-1}K^{-1}}\times300\ \mathrm{K}\times\ln(2\times10^{13})$

Answer

(c) $-8.314\ \mathrm{J\,mol^{-1}K^{-1}}\times300\ \mathrm{K}\times\ln(2\times10^{13})$

$\Delta_rG^\ominus=-RT\ln K_c$ . A very large $K_c$ means a large negative $\Delta_rG^\ominus$ (strongly spontaneous). Options with a positive sign or with a different number inside the logarithm do not match.

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2020 · NEET-UG (India) · 4 marks

For the reaction $2\mathrm{Cl(g)} \rightarrow \mathrm{Cl_2(g)}$ , the correct option is:
(a) $\Delta_r H < 0$ and $\Delta_r S > 0$ (b) $\Delta_r H < 0$ and $\Delta_r S < 0$ (c) $\Delta_r H > 0$ and $\Delta_r S > 0$ (d) $\Delta_r H > 0$ and $\Delta_r S < 0$

Answer

(b) $\Delta_rH<0$ and $\Delta_rS<0$

Forming a Cl-Cl bond from atoms releases energy, so $\Delta_rH<0$ . Two moles of gas become one, so disorder falls and $\Delta_rS<0$ .

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2020 · NEET-UG (India) · 4 marks

Find out the solubility of $\mathrm{Ni(OH)_2}$ in 0.1 M NaOH. Given that the solubility product of $\mathrm{Ni(OH)_2}$ is $2\times10^{-15}$ .
(a) $1\times10^{-13}\ \mathrm{M}$ (b) $1\times10^{8}\ \mathrm{M}$ (c) $2\times10^{-13}\ \mathrm{M}$ (d) $2\times10^{-8}\ \mathrm{M}$

Answer

(c) $2\times10^{-13}$ M

NaOH gives $[\mathrm{OH^-}]=0.1$ M, which suppresses the dissolution of $\mathrm{Ni(OH)_2}$ (common ion effect). With solubility $s$ , $K_{sp}=s(0.1)^2$ , so $s=\dfrac{2\times10^{-15}}{10^{-2}}=2\times10^{-13}$ M. The $\mathrm{OH^-}$ from the salt itself is negligible.

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2020 · NEET-UG (India) · 4 marks

On electrolysis of dil. sulphuric acid using platinum (Pt) electrode, the product obtained at anode will be:
(a) $\mathrm{H_2S}$ gas (b) $\mathrm{SO_2}$ gas (c) Hydrogen gas (d) Oxygen gas

Answer

(d) Oxygen gas

With inert Pt electrodes in dilute $\mathrm{H_2SO_4}$ , water is oxidised at the anode: $2\mathrm{H_2O}\rightarrow\mathrm{O_2}+4\mathrm{H^+}+4e^-$ . Sulphate is not oxidised in dilute solution. Hydrogen is released at the cathode, not the anode.

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2020 · NEET-UG (India) · 4 marks

The number of Faradays (F) required to produce 20 g of calcium from molten $\mathrm{CaCl_2}$ (Atomic mass of Ca = 40 g $\mathrm{mol^{-1}}$ ) is:
(a) 3 (b) 4 (c) 1 (d) 2

Answer

(c) 1

$\mathrm{Ca^{2+}} + 2e^- \rightarrow \mathrm{Ca}$ , so 1 mol (40 g) of Ca needs 2 F. 20 g of Ca is $20/40 = 0.5$ mol, so charge $= 0.5 \times 2 = 1$ F.

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2020 · NEET-UG (India) · 4 marks

What is the change in oxidation number of carbon in the following reaction? $\mathrm{CH_4(g)} + 4\mathrm{Cl_2(g)} \rightarrow \mathrm{CCl_4(l)} + 4\mathrm{HCl(g)}$
(a) $-4$ to $+4$ (b) 0 to $-4$ (c) $+4$ to $+4$ (d) 0 to $+4$

Answer

(a) $-4$ to $+4$

In $\mathrm{CH_4}$ , carbon is bonded to four H (less electronegative), so $x + 4(+1) = 0$ gives $x = -4$ . In $\mathrm{CCl_4}$ , Cl is more electronegative, so $x + 4(-1) = 0$ gives $x = +4$ .

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2020 · NEET-UG (India) · 4 marks

An increase in the concentration of the reactants of a reaction leads to change in:
(a) Threshold energy (b) Collision frequency (c) Activation energy (d) Heat of reaction

Answer

(b) Collision frequency

Higher concentration puts more molecules per unit volume, so collisions per second increase and the rate rises. Threshold energy, activation energy and heat of reaction depend on the nature of reactants and the reaction path (and temperature or catalyst), not on concentration.

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2020 · NEET-UG (India) · 4 marks

The mixture which shows positive deviation from Raoult's law is:
(a) Acetone + Chloroform (b) Chloroethane + Bromoethane (c) Ethanol + Acetone (d) Benzene + Toluene

Answer

(c) Ethanol + Acetone

Ethanol is strongly hydrogen-bonded; adding acetone breaks some of these bonds, so the A-B interactions are weaker than A-A and B-B. Vapour pressure is higher than Raoult predicts (positive deviation). Acetone + chloroform forms H-bonds and shows negative deviation; chloroethane + bromoethane and benzene + toluene are nearly ideal.

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2020 · NEET-UG (India) · 4 marks

The freezing point depression constant ( $K_f$ ) of benzene is 5.12 K kg $\mathrm{mol^{-1}}$ . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places):
(a) 0.40 K (b) 0.60 K (c) 0.20 K (d) 0.80 K

Answer

(a) 0.40 K

$\Delta T_f = K_f \times m = 5.12 \times 0.078 = 0.399 \approx 0.40$ K. The solute is a non-electrolyte, so the van't Hoff factor $i = 1$ and no correction is needed.

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2021 · NEET-UG (India) · 4 marks

Which of the following reactions is the metal displacement reaction? Choose the right option.
(a) $\mathrm{Fe + 2HCl \longrightarrow FeCl_2 + H_2\uparrow}$ (b) $\mathrm{2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2\uparrow}$ (c) $\mathrm{2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2}$ (d) $\mathrm{Cr_2O_3 + 2Al \xrightarrow{\Delta} Al_2O_3 + 2Cr}$

Answer

(d) $\mathrm{Cr_2O_3 + 2Al \xrightarrow{\Delta} Al_2O_3 + 2Cr}$

In a metal displacement reaction a more reactive metal displaces a less reactive metal from its compound. Here Al displaces Cr from $\mathrm{Cr_2O_3}$ . Option (a) is a metal displacing hydrogen (not a metal), and (b) and (c) are thermal decompositions.

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2021 · NEET-UG (India) · 4 marks

The $pK_b$ of dimethylamine and $pK_a$ of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethyl ammonium acetate solution is:
(a) 7.75 (b) 6.25 (c) 8.50 (d) 5.50

Answer

(a) 7.75

For a salt of a weak acid and a weak base, $\mathrm{pH}=7+\tfrac12(pK_a-pK_b)$ . $\mathrm{pH}=7+\tfrac12(4.77-3.27)=7+0.75=7.75$ . The solution is slightly basic because the base is a little stronger than the acid.

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2021 · NEET-UG (India) · 4 marks

The molar conductance of NaCl, HCl and $\mathrm{CH_3COONa}$ at infinite dilution are 126.45, 426.16 and 91.0 $\mathrm{S\,cm^2\,mol^{-1}}$ respectively. The molar conductance of $\mathrm{CH_3COOH}$ at infinite dilution is. Choose the right option for your answer.
(a) 698.28 $\mathrm{S\,cm^2\,mol^{-1}}$ (b) 540.48 $\mathrm{S\,cm^2\,mol^{-1}}$ (c) 201.28 $\mathrm{S\,cm^2\,mol^{-1}}$ (d) 390.71 $\mathrm{S\,cm^2\,mol^{-1}}$

Answer

(d) $390.71\ \mathrm{S\,cm^2\,mol^{-1}}$

By Kohlrausch law, $\Lambda^\circ(\mathrm{CH_3COOH})=\Lambda^\circ(\mathrm{HCl})+\Lambda^\circ(\mathrm{CH_3COONa})-\Lambda^\circ(\mathrm{NaCl})$ . $=426.16+91.0-126.45=390.71\ \mathrm{S\,cm^2\,mol^{-1}}$ . The $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ contributions cancel.

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2021 · NEET-UG (India) · 4 marks

Which one among the following is the correct option for right relationship between $C_P$ and $C_V$ for one mole of ideal gas?
(a) $C_P = RC_V$ (b) $C_V = RC_P$ (c) $C_P + C_V = R$ (d) $C_P - C_V = R$

Answer

(d) $C_P - C_V = R$

For one mole of an ideal gas, $H=U+RT$ , so $\dfrac{dH}{dT}=\dfrac{dU}{dT}+R$ , which gives $C_P-C_V=R$ . $C_P$ is larger because at constant pressure some heat also goes into expansion work.

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2021 · NEET-UG (India) · 4 marks

The following solutions were prepared by dissolving 10 g glucose ( $\mathrm{C_6H_{12}O_6}$ ) in 250 mL of water (P1), 10 g of urea ( $\mathrm{CH_4N_2O}$ ) in 250 mL of water (P2) and 10 g of sucrose ( $\mathrm{C_{12}H_{22}O_{11}}$ ) in 250 mL of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is.
(a) P2 > P3 > P1 (b) P3 > P1 > P2 (c) P2 > P1 > P3 (d) P1 > P2 > P3

Answer

(c) P2 > P1 > P3

For solutions of the same volume and temperature, $\pi=CRT\propto$ moles of solute (all are non-electrolytes). Glucose $=10/180=0.056$ mol, urea $=10/60=0.167$ mol, sucrose $=10/342=0.029$ mol, so urea > glucose > sucrose.

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2021 · NEET-UG (India) · 4 marks

Choose the correct option for the total pressure (in atm.) in a mixture of 4 g $\mathrm{O_2}$ and 2 g $\mathrm{H_2}$ confined in a total volume of one litre at $0^\circ\mathrm{C}$ is:
(a) 25.18 (b) 26.02 (c) 2.518 (d) 2.602

Answer

(a) 25.18

$n(\mathrm{O_2})=4/32=0.125$ mol and $n(\mathrm{H_2})=2/2=1$ mol, so $n_{total}=1.125$ mol. $P=\dfrac{nRT}{V}=\dfrac{1.125\times0.082\times273}{1}\approx25.18$ atm. The value 2.518 is off by a factor of ten.

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2021 · NEET-UG (India) · 4 marks

The correct option for the value of vapour pressure of a solution at $45^\circ\mathrm{C}$ with benzene to octane in molar ratio 3 : 2 is: [At $45^\circ\mathrm{C}$ vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal gas]
(a) 336 mm of Hg (b) 350 mm of Hg (c) 160 mm of Hg (d) 168 mm of Hg

Answer

(a) 336 mm of Hg

By Raoult law, $P=x_B P_B^\circ+x_O P_O^\circ$ . The molar ratio 3:2 gives $x_B=0.6$ and $x_O=0.4$ . $P=0.6\times280+0.4\times420=168+168=336$ mm Hg.

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2021 · NEET-UG (India) · 4 marks

For irreversible expansion of an ideal gas under isothermal condition, the correct option is:
(a) $\Delta U = 0,\ \Delta S_{total} \neq 0$ (b) $\Delta U \neq 0,\ \Delta S_{total} = 0$ (c) $\Delta U = 0,\ \Delta S_{total} = 0$ (d) $\Delta U \neq 0,\ \Delta S_{total} \neq 0$

Answer

(a) $\Delta U = 0,\ \Delta S_{total} \neq 0$

For an ideal gas at constant temperature, internal energy depends only on $T$ , so $\Delta U=0$ . An irreversible process is spontaneous, so the entropy of the universe increases and $\Delta S_{total}>0$ . Only a reversible process has $\Delta S_{total}=0$ .

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2021 · NEET-UG (India) · 4 marks

The molar conductivity of 0.007 M acetic acid is 20 $\mathrm{S\,cm^2\,mol^{-1}}$ . What is the dissociation constant of acetic acid? Choose the correct option. $\lambda^\circ_{\mathrm{H^+}} = 350\ \mathrm{S\,cm^2\,mol^{-1}}$ , $\lambda^\circ_{\mathrm{CH_3COO^-}} = 50\ \mathrm{S\,cm^2\,mol^{-1}}$
(a) $1.75\times10^{-5}\ \mathrm{mol\,L^{-1}}$ (b) $2.50\times10^{-5}\ \mathrm{mol\,L^{-1}}$ (c) $1.75\times10^{-4}\ \mathrm{mol\,L^{-1}}$ (d) $2.50\times10^{-4}\ \mathrm{mol\,L^{-1}}$

Answer

(a) $1.75\times10^{-5}\ \mathrm{mol\,L^{-1}}$

$\Lambda^\circ=350+50=400$ , so $\alpha=\Lambda/\Lambda^\circ=20/400=0.05$ . $K_a=\dfrac{C\alpha^2}{1-\alpha}\approx C\alpha^2=0.007\times(0.05)^2=1.75\times10^{-5}$ mol/L.

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2021 · NEET-UG (India) · 4 marks

The slope of Arrhenius Plot ( $\ln k$ vs $1/T$ ) of first order reaction is $-5\times10^{3}$ K. The value of $E_a$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$ ]
(a) 166 kJ $\mathrm{mol^{-1}}$ (b) -83 kJ $\mathrm{mol^{-1}}$ (c) 41.5 kJ $\mathrm{mol^{-1}}$ (d) 83.0 kJ $\mathrm{mol^{-1}}$

Answer

(c) 41.5 kJ $\mathrm{mol^{-1}}$

From $\ln k=\ln A-\dfrac{E_a}{RT}$ , the slope of $\ln k$ vs $1/T$ is $-E_a/R$ . $E_a=5\times10^3\times8.314=41570$ J/mol $\approx41.5$ kJ/mol. A negative $E_a$ is not possible here.

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2022 · NEET-UG (India) · 4 marks

Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here $p$ = total pressure of gaseous mixture
(a) $p = p_1 + p_2 + p_3$ (b) $p = n_1\frac{RT}{V} + n_2\frac{RT}{V} + n_3\frac{RT}{V}$ (c) $p_i = \chi_i\, p$ , where $p_i$ = partial pressure of $i$ th gas; $\chi_i$ = mole fraction of $i$ th gas in gaseous mixture (d) $p_i = \chi_i\, p_i^{\circ}$ , where $\chi_i$ = mole fraction of $i$ th gas in gaseous mixture, $p_i^{\circ}$ = pressure of $i$ th gas in pure state

Answer

(d) $p_i = \chi_i\, p_i^{\circ}$

Dalton law gives $p=p_1+p_2+p_3$ , and since $p_i=n_iRT/V$ this is also (b). Using $p_i=\chi_i p$ is also correct. The relation $p_i=\chi_i p_i^\circ$ is Raoult law for a solution, not Dalton law.

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2022 · NEET-UG (India) · 4 marks

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given $pK_a$ of $\mathrm{CH_3COOH} = 4.57$ ]
(a) 5.57 (b) 3.57 (c) 4.57 (d) 2.57

Answer

(a) 5.57

This is an acidic buffer: $\mathrm{pH}=pK_a+\log\dfrac{[\text{salt}]}{[\text{acid}]}$ . Equal volumes are mixed, so the ratio stays $0.10/0.01=10$ , giving $\mathrm{pH}=4.57+\log10=5.57$ .

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2022 · NEET-UG (India) · 4 marks

Given below are half cell reactions: $\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$ , $E^\circ_{\mathrm{Mn^{2+}/MnO_4^-}} = -1.510\ \mathrm{V}$ $\frac{1}{2}\mathrm{O_2} + 2\mathrm{H^+} + 2e^- \rightarrow \mathrm{H_2O}$ , $E^\circ_{\mathrm{O_2/H_2O}} = +1.223\ \mathrm{V}$ Will the permanganate ion, $\mathrm{MnO_4^-}$ liberate $\mathrm{O_2}$ from water in the presence of an acid?
(a) Yes, because $E^\circ_{cell} = +0.287\ \mathrm{V}$ (b) No, because $E^\circ_{cell} = -0.287\ \mathrm{V}$ (c) Yes, because $E^\circ_{cell} = +2.733\ \mathrm{V}$ (d) No, because $E^\circ_{cell} = -2.733\ \mathrm{V}$

Answer

(b) No, because $E^\circ_{cell} = -0.287\ \mathrm{V}$

Here $\mathrm{MnO_4^-}$ is reduced and water is oxidised to $\mathrm{O_2}$ . With the potentials as given, $E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.223 - 1.510 = -0.287\ \mathrm{V}$ (using the question's sign convention for the pair). A negative $E^\circ_{cell}$ means the reaction is non-spontaneous, so $\mathrm{O_2}$ is not liberated.

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2022 · NEET-UG (India) · 4 marks

At 298 K, the standard electrode potentials of $\mathrm{Cu^{2+}/Cu}$ , $\mathrm{Zn^{2+}/Zn}$ , $\mathrm{Fe^{2+}/Fe}$ and $\mathrm{Ag^+/Ag}$ are 0.34 V, $-0.76$ V, $-0.44$ V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?
(a) $\mathrm{CuSO_4(aq) + Zn(s) \rightarrow ZnSO_4(aq) + Cu(s)}$ (b) $\mathrm{CuSO_4(aq) + Fe(s) \rightarrow FeSO_4(aq) + Cu(s)}$ (c) $\mathrm{FeSO_4(aq) + Zn(s) \rightarrow ZnSO_4(aq) + Fe(s)}$ (d) $\mathrm{2CuSO_4(aq) + 2Ag(s) \rightarrow 2Cu(s) + Ag_2SO_4(aq)}$

Answer

(d) $\mathrm{2CuSO_4(aq) + 2Ag(s) \rightarrow 2Cu(s) + Ag_2SO_4(aq)}$

A metal displaces another from its salt only if its $E^\circ$ is lower (more negative) than that of the ion being reduced. Here Ag ( $+0.80$ V) would have to reduce $\mathrm{Cu^{2+}}$ ( $+0.34$ V): $E^\circ_{cell} = 0.34 - 0.80 = -0.46$ V, non-spontaneous. In (a), (b), (c) Zn or Fe ( $-0.76$ , $-0.44$ V) lie below the ion displaced, giving positive $E^\circ_{cell}$ , so they occur.

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2022 · NEET-UG (India) · 4 marks

In one molal solution that contains 0.5 mole of a solute, there is
(a) 500 mL of solvent (b) 500 g of solvent (c) 100 mL of solvent (d) 1000 g of solvent

Answer

(b) 500 g of solvent

Molality $= \dfrac{\text{moles of solute}}{\text{kg of solvent}}$ . For $1\ \mathrm{m}$ with $0.5$ mol: $1 = \dfrac{0.5}{w}$ , so $w = 0.5$ kg $= 500$ g of solvent. Molality is defined by mass of solvent, so the volume options are wrong.

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2022 · NEET-UG (India) · 4 marks

A 10.0 L flask contains 64 g of oxygen at $27^\circ\mathrm{C}$ . (Assume $\mathrm{O_2}$ gas is behaving ideally). The pressure inside the flask in bar is (Given $R = 0.0831\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}}$ )
(a) 2.5 (b) 498.6 (c) 49.8 (d) 4.9

Answer

(d) 4.9

$n = 64/32 = 2$ mol, $T = 300$ K. $P = \dfrac{nRT}{V} = \dfrac{2 \times 0.0831 \times 300}{10} = 4.986 \approx 4.9$ bar. Option (a) would result from using $T=273$ or a wrong value of $R$ .

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2022 · NEET-UG (India) · 4 marks

For a first order reaction A $\rightarrow$ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in $\mathrm{min^{-1}}$ is
(a) 1.3818 (b) 0.9212 (c) 0.4606 (d) 0.2303

Answer

(b) 0.9212

For a first-order reaction $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]} = \dfrac{2.303}{5}\log\dfrac{0.1}{0.001} = \dfrac{2.303}{5}\times 2 = 0.9212\ \mathrm{min^{-1}}$ .

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2022 · NEET-UG (India) · 4 marks

$\mathrm{3O_2(g) \rightleftharpoons 2O_3(g)}$ . For the above reaction at 298 K, $K_c$ is found to be $3.0\times10^{-59}$ . If the conc. of $\mathrm{O_2}$ at equilibrium is 0.040 M then concentration of $\mathrm{O_3}$ in M is.
(a) $4.38\times10^{-32}$ (b) $1.9\times10^{-63}$ (c) $2.4\times10^{31}$ (d) $1.2\times10^{21}$

Answer

(a) $4.38\times10^{-32}$

$K_c = \dfrac{[\mathrm{O_3}]^2}{[\mathrm{O_2}]^3}$ , so $[\mathrm{O_3}]^2 = 3.0\times10^{-59}\times(0.040)^3 = 3.0\times10^{-59}\times6.4\times10^{-5} = 1.92\times10^{-63}$ . Hence $[\mathrm{O_3}] = \sqrt{1.92\times10^{-63}} = 4.38\times10^{-32}$ M.

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2022 · NEET-UG (India) · 4 marks

Find the emf of the cell $\mathrm{Ni(s) + 2Ag^+(0.001\ M) \rightarrow Ni^{2+}(0.001\ M) + 2Ag(s)}$ (Given that $E^\circ_{cell} = 1.05\ \mathrm{V}$ , $\frac{2.303RT}{F} = 0.059$ at 298 K)
(a) 1.0385 V (b) 1.385 V (c) 0.9615 V (d) 1.05 V

Answer

(c) 0.9615 V

Nernst equation with $n=2$ : $E = E^\circ - \dfrac{0.059}{2}\log\dfrac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^+}]^2} = 1.05 - 0.0295\log\dfrac{10^{-3}}{10^{-6}} = 1.05 - 0.0295\times3 = 0.9615$ V.

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2023 · NEET-UG (India) · 4 marks

Assertion A: A reaction can have zero activation energy. Reason R: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy. In the light of the above statements, choose the correct answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true and R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Answer

(d) A is false but R is true

A reaction cannot have zero activation energy: with $E_a = 0$ the Arrhenius equation gives $k = A$ , meaning every collision would be effective, which does not happen. Reason R correctly states the definition of activation energy (threshold energy minus average energy of reactants).

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2023 · NEET-UG (India) · 4 marks

The conductivity of centimolar solution of KCl at $25^\circ\mathrm{C}$ is $0.210\ \mathrm{ohm^{-1}\,cm^{-1}}$ and the resistance of the cell containing the solution at $25^\circ\mathrm{C}$ is 60 ohm. The value of cell constant is
(a) $1.34\ \mathrm{cm^{-1}}$ (b) $3.28\ \mathrm{cm^{-1}}$ (c) $1.26\ \mathrm{cm^{-1}}$ (d) $3.34\ \mathrm{cm^{-1}}$

Answer

(c) $1.26\ \mathrm{cm^{-1}}$

Cell constant $= \kappa \times R = 0.210 \times 60$ . This evaluates to $12.6\ \mathrm{cm^{-1}}$ , but the NTA key lists $1.26\ \mathrm{cm^{-1}}$ , the only option with the digits 126, so (c) is the intended answer (the question data is known to be inconsistent by a factor of ten).

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2023 · NEET-UG (India) · 4 marks

For a certain reaction, the rate $= k[\mathrm{A}]^2[\mathrm{B}]$ . When the initial concentration of A is tripled keeping concentration of B constant, the initial rate would
(a) Decrease by a factor of nine (b) Increase by a factor of six (c) Increase by a factor of nine (d) Increase by a factor of three

Answer

(c) Increase by a factor of nine

Rate $=k[A]^2[B]$ is second order in A. Tripling A with B constant gives rate $\propto (3)^2 = 9$ , so the rate increases nine times.

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2023 · NEET-UG (India) · 4 marks

Amongst the given options which of the following molecules/ion acts as a Lewis acid?
(a) $\mathrm{NH_3}$ (b) $\mathrm{H_2O}$ (c) $\mathrm{BF_3}$ (d) $\mathrm{OH^-}$

Answer

(c) $\mathrm{BF_3}$

A Lewis acid accepts an electron pair. In $\mathrm{BF_3}$ boron has only six electrons (incomplete octet) and an empty p orbital. $\mathrm{NH_3}$ , $\mathrm{H_2O}$ and $\mathrm{OH^-}$ all have lone pairs to donate, so they are Lewis bases.

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2023 · NEET-UG (India) · 4 marks

Intermolecular forces are forces of attraction and repulsion between interacting particles that will include A. dipole - dipole forces B. dipole - induced dipole forces C. hydrogen bonding D. covalent bonding E. dispersion forces Choose the most appropriate answer from the options given below :
(a) B, C, D, E are correct (b) A, B, C, D are correct (c) A, B, C, E are correct (d) A, C, D, E are correct

Answer

(c) A, B, C, E are correct

Intermolecular forces include dipole-dipole, dipole-induced dipole, hydrogen bonding and dispersion (London) forces. Covalent bonding (D) is an intramolecular force holding atoms within a molecule, so it is excluded.

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2023 · NEET-UG (India) · 4 marks

Assertion A: In equation $\Delta_rG = -nFE_{cell}$ , value of $\Delta_rG$ depends on $n$ . Reason R: $E_{cell}$ is an intensive property and $\Delta_rG$ is an extensive property. In the light of the above statements, choose the correct answer from the options given below
(a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true and R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Answer

(a) Both A and R are true and R is the correct explanation of A

In $\Delta_rG=-nFE_{cell}$ , $\Delta_rG$ is directly proportional to $n$ , so A is true. $E_{cell}$ is intensive (independent of amount) while $\Delta_rG$ is extensive, which is exactly why $\Delta_rG$ changes with $n$ ; R explains A.

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2023 · NEET-UG (India) · 4 marks

Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
(a) $\Delta H = \Delta U - \Delta n_gRT$ (b) $\Delta H = \Delta U + \Delta n_gRT$ (c) $\Delta H - \Delta U = -\Delta nRT$ (d) $\Delta H + \Delta U = \Delta nR$

Answer

(b) $\Delta H = \Delta U + \Delta n_gRT$

From $H=U+PV$ and $PV=nRT$ for gases at constant temperature, $\Delta H=\Delta U+\Delta n_g RT$ . The sign is positive; option (a) has the sign wrong.

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2023 · NEET-UG (India) · 4 marks

The equilibrium concentrations of the species in the reaction $\mathrm{A + B \rightleftharpoons C + D}$ are 2, 3, 10 and 6 $\mathrm{mol\,L^{-1}}$ , respectively at 300 K. $\Delta G^\circ$ for the reaction is ( $R = 2$ cal/mol K)
(a) 1372.60 cal (b) -137.26 cal (c) -1381.80 cal (d) -13.73 cal

Answer

(c) -1381.80 cal

$K=\dfrac{[C][D]}{[A][B]}=\dfrac{10\times6}{2\times3}=10$ . $\Delta G^\circ=-2.303RT\log K=-2.303\times2\times300\times1=-1381.8$ cal. The sign is negative because $K>1$ .

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2024 · NEET-UG (India) · 4 marks

For the reaction $\mathrm{2A \rightleftharpoons B + C}$ , $K_c = 4\times10^{-3}$ . At a given time, the composition of reaction mixture is: $[\mathrm{A}] = [\mathrm{B}] = [\mathrm{C}] = 4\times10^{-3}$ . Then, which of the following is correct?
(a) Reaction has a tendency to go in forward direction. (b) Reaction has a tendency to go in backward direction. (c) Reaction has gone to completion in forward direction. (d) Reaction is at equilibrium.

Answer

(b) Reaction has a tendency to go in backward direction.

$Q_c=\dfrac{[B][C]}{[A]^2}=\dfrac{(4\times10^{-3})^2}{(4\times10^{-3})^2}=1$ . Since $Q_c=1>K_c=4\times10^{-3}$ , products exceed the equilibrium amount, so the reaction shifts backward.

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2024 · NEET-UG (India) · 4 marks

Which reaction is NOT a redox reaction ?
(a) $\mathrm{2KClO_3 + I_2 \rightarrow 2KIO_3 + Cl_2}$ (b) $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$ (c) $\mathrm{BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl}$ (d) $\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}$

Answer

(c) $\mathrm{BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl}$

This is a double displacement (precipitation) reaction in which no oxidation numbers change. The others involve changes in oxidation state: Cl and I in (a), H and Cl in (b), Zn and Cu in (d).

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2024 · NEET-UG (India) · 4 marks

In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. $\mathrm{2NaHCO_3(s) \rightarrow Na_2CO_3(s) + CO_2(g) + H_2O(g)}$ D. $\mathrm{Cl_2(g) \rightarrow 2Cl(g)}$ Choose the correct answer from the options given below:
(a) A, B and D (b) A, C and D (c) C and D (d) A and C

Answer

(b) A, C and D

Entropy rises when disorder increases: evaporation (A), solid decomposing to give gases (C), and dissociation of $\mathrm{Cl_2}$ into atoms (D). Cooling a crystalline solid towards 0 K (B) lowers entropy.

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2024 · NEET-UG (India) · 4 marks

Activation energy of any chemical reaction can be calculated if one knows the value of
(a) Probability of collision. (b) Orientation of reactant molecules during collision. (c) rate constant at two different temperatures. (d) rate constant at standard temperature.

Answer

(c) rate constant at two different temperatures.

From $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ , $E_a$ can be found if $k$ is known at two temperatures. A single $k$ at one temperature is not enough, since the frequency factor is also unknown.

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2024 · NEET-UG (India) · 4 marks

In which of the following equilibria, $K_p$ and $K_c$ are NOT equal?
(a) $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ (b) $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ (c) $\mathrm{2BrCl(g) \rightleftharpoons Br_2(g) + Cl_2(g)}$ (d) $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$

Answer

(d) $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$

$K_p=K_c(RT)^{\Delta n_g}$ , so they differ only if $\Delta n_g\neq0$ . For (a), (b) and (c) $\Delta n_g=0$ . For $\mathrm{PCl_5}$ , $\Delta n_g=2-1=1$ , so $K_p\neq K_c$ .

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2024 · NEET-UG (India) · 4 marks

Match List-I (Conversion) with List-II (No. of Faraday required). List-I A. 1 mol of $\mathrm{H_2O}$ to $\mathrm{O_2}$ B. 1 mol of $\mathrm{MnO_4^-}$ to $\mathrm{Mn^{2+}}$ C. 1.5 mol of Ca from molten $\mathrm{CaCl_2}$ D. 1 mol of FeO to $\mathrm{Fe_2O_3}$ List-II I. 3F II. 2F III. 1F IV. 5F
(a) A-III, B-IV, C-I, D-II (b) A-II, B-III, C-I, D-IV (c) A-III, B-IV, C-II, D-I (d) A-II, B-IV, C-I, D-III

Answer

(d) A-II, B-IV, C-I, D-III

$\mathrm{H_2O}\to\frac12\mathrm{O_2}$ takes 2 e per mole, so 2F. $\mathrm{MnO_4^-}\to\mathrm{Mn^{2+}}$ takes 5 e, so 5F. 1.5 mol Ca needs $1.5\times2=3$ F. $\mathrm{FeO}\to\mathrm{Fe_2O_3}$ is $\mathrm{Fe^{2+}}\to\mathrm{Fe^{3+}}$ , 1 e per mole, so 1F.

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2024 · NEET-UG (India) · 4 marks

Match List-I (Process) with List-II (Conditions). List-I A. Isothermal process B. Isochoric process C. Isobaric process D. Adiabatic process List-II I. No heat exchange II. Carried out at constant temp. III. Carried out at constant volume IV. Carried out at constant pressure
(a) A-IV, B-II, C-III, D-I (b) A-I, B-II, C-III, D-IV (c) A-II, B-III, C-IV, D-I (d) A-IV, B-III, C-II, D-I

Answer

(c) A-II, B-III, C-IV, D-I

Isothermal means constant temperature, isochoric constant volume, isobaric constant pressure. Adiabatic means no heat exchange with the surroundings.

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2024 · NEET-UG (India) · 4 marks

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is: (Given: Molar mass of Cu: 63 g $\mathrm{mol^{-1}}$ , 1 F = 96487 C)
(a) 0.315 g (b) 31.5 g (c) 0.0315 g (d) 3.15 g

Answer

(a) 0.315 g

$Q=It=9.6487\times100=964.87$ C, which is $964.87/96487=0.01$ F. $\mathrm{Cu^{2+}+2e^-\rightarrow Cu}$ , so equivalent mass $=63/2=31.5$ g. Mass $=0.01\times31.5=0.315$ g.

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2024 · NEET-UG (India) · 4 marks

Consider the following reactions in a sealed vessel at equilibrium with concentrations of $\mathrm{N_2} = 3.0\times10^{-3}$ M, $\mathrm{O_2} = 4.2\times10^{-3}$ M and $\mathrm{NO} = 2.8\times10^{-3}$ M: $\mathrm{2NO(g) \rightleftharpoons N_2(g) + O_2(g)}$ . If 0.1 mol $\mathrm{L^{-1}}$ of NO(g) is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of NO(g) at equilibrium?
(a) 0.0889 (b) 0.8889 (c) 0.717 (d) 0.00889

Answer

(c) $0.717$

$K_c=\dfrac{[\mathrm{N_2}][\mathrm{O_2}]}{[\mathrm{NO}]^2}=\dfrac{(3.0\times10^{-3})(4.2\times10^{-3})}{(2.8\times10^{-3})^2}=1.607$ . For $0.1$ M NO with $2x$ dissociated: $\dfrac{x^2}{(0.1-2x)^2}=1.607$ , so $\dfrac{x}{0.1-2x}=1.268$ and $x=0.0359$ . $\alpha=\dfrac{2x}{0.1}=0.717$ . The 0.0889 options come from forgetting the factor 2 or inverting $K_c$ .

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2024 · NEET-UG (India) · 4 marks

The work done during reversible isothermal expansion of one mole of hydrogen gas at $25^\circ\mathrm{C}$ from pressure of 20 atmosphere to 10 atmosphere is: (Given $R = 2.0\ \mathrm{cal\,K^{-1}\,mol^{-1}}$ )
(a) -413.14 calories (b) 413.14 calories (c) 100 calories (d) 0 calories

Answer

(a) $-413.14$ calories

For a reversible isothermal expansion, $W=-nRT\ln\dfrac{P_1}{P_2}=-2.303\,nRT\log\dfrac{P_1}{P_2}$ . $W=-2.303\times1\times2\times298\times\log2=-2.303\times596\times0.301\approx-413.14$ cal. The sign is negative because the gas does work on the surroundings (work done by the system). Option (b) has the wrong sign.

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2024 · NEET-UG (India) · 4 marks

The rate of a reaction quadruples when temperature changes from $27^\circ\mathrm{C}$ to $57^\circ\mathrm{C}$ . Calculate the energy of activation. Given $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$ , $\log 4 = 0.6021$
(a) 380.4 kJ/mol (b) 3.80 kJ/mol (c) 3804 kJ/mol (d) 38.04 kJ/mol

Answer

(d) $38.04$ kJ/mol

$\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right)$ with $T_1=300$ K and $T_2=330$ K. $E_a=\dfrac{2.303\times8.314\times0.6021\times300\times330}{30}\approx3.804\times10^4$ J $=38.04$ kJ/mol.

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2024 · NEET-UG (India) · 4 marks

The plot of osmotic pressure ( $\Pi$ ) vs concentration ( $\mathrm{mol\,L^{-1}}$ ) for a solution gives a straight line with slope 25.73 $\mathrm{L\,bar\,mol^{-1}}$ . The temperature at which the osmotic pressure measurement is done is: (Use $R = 0.083\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$ )
(a) $310^\circ\mathrm{C}$ (b) $25.73^\circ\mathrm{C}$ (c) $12.05^\circ\mathrm{C}$ (d) $37^\circ\mathrm{C}$

Answer

(d) $37^\circ\mathrm{C}$

$\Pi=CRT$ , so the slope of $\Pi$ against $C$ equals $RT$ . $T=\dfrac{25.73}{0.083}=310$ K $=37^\circ$ C. The value $310^\circ$ C in option (a) is a tempting trap because it forgets to convert from kelvin to Celsius.

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2025 · NEET-UG (India) · 4 marks

If the rate constant of a reaction is $0.03\ \mathrm{s^{-1}}$ , how much time does it take for 7.2 $\mathrm{mol\,L^{-1}}$ concentration of the reactant to get reduced to 0.9 $\mathrm{mol\,L^{-1}}$ ? (Given: $\log 2 = 0.301$ )
(a) 210 s (b) 21.0 s (c) 69.3 s (d) 23.1 s

Answer

(c) $69.3$ s

The concentration falls from $7.2$ to $0.9$ , a factor of $8=2^3$ , so this is three half-lives. $t=\dfrac{2.303}{k}\log\dfrac{7.2}{0.9}=\dfrac{2.303\times0.903}{0.03}\approx69.3$ s. The same result comes from $3\times t_{1/2}=3\times\dfrac{0.693}{0.03}=69.3$ s.

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2025 · NEET-UG (India) · 4 marks

Which of the following aqueous solution will exhibit highest boiling point?
(a) 0.01 M $\mathrm{Na_2SO_4}$ (b) 0.015 M $\mathrm{C_6H_{12}O_6}$ (c) 0.01 M Urea (d) 0.01 M $\mathrm{KNO_3}$

Answer

(a) $0.01$ M $\mathrm{Na_2SO_4}$

Elevation of boiling point is a colligative property, proportional to $i\times C$ . $\mathrm{Na_2SO_4}$ : $i=3$ , so $i C=0.03$ . Glucose: $0.015$ . Urea: $0.01$ . $\mathrm{KNO_3}$ : $i=2$ , so $0.02$ . $\mathrm{Na_2SO_4}$ has the largest value and so the highest boiling point.

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2025 · NEET-UG (India) · 4 marks

The standard heat of formation, in kcal/mol of $\mathrm{Ba^{2+}}$ is: [Given: standard heat of formation of $\mathrm{SO_4^{2-}}$ ion (aq) $= -216$ kcal/mol, standard heat of crystallisation of $\mathrm{BaSO_4(s)} = -4.5$ kcal/mol, standard heat of formation of $\mathrm{BaSO_4(s)} = -349$ kcal/mol]
(a) +133.0 (b) +220.5 (c) -128.5 (d) -133.0

Answer

(c) $-128.5$

For $\mathrm{Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)}$ , the enthalpy of crystallisation is $-4.5$ kcal. $\Delta H_f(\mathrm{BaSO_4})=\Delta H_f(\mathrm{Ba^{2+}})+\Delta H_f(\mathrm{SO_4^{2-}})+\Delta H_{cryst}$ . $-349=x-216-4.5$ , so $x=-128.5$ kcal/mol.

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2025 · NEET-UG (India) · 4 marks

Match List-I (Example) with List-II (Type of Solution). List-I A. Humidity B. Alloys C. Amalgams D. Smoke List-II I. Solid in solid II. Liquid in gas III. Solid in gas IV. Liquid in solid
(a) A-III, B-I, C-IV, D-II (b) A-III, B-II, C-I, D-IV (c) A-II, B-IV, C-I, D-III (d) A-II, B-I, C-IV, D-III

Answer

(d) A-II, B-I, C-IV, D-III

Humidity is liquid (water) dispersed in gas (air). Alloys are solid in solid. Amalgams are liquid (mercury) in solid. Smoke is solid particles in gas.

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2025 · NEET-UG (India) · 4 marks

For the reaction $\mathrm{A(g) \rightleftharpoons 2B(g)}$ , the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given: $R = 0.0831\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$ ] $K_p$ for the reaction at 1000 K is
(a) 0.033 (b) 0.021 (c) 83.1 (d) $2.077\times10^{5}$

Answer

(a) $0.033$

$K_c=\dfrac{k_f}{k_b}=\dfrac{1}{2500}=4\times10^{-4}$ . $\Delta n_g=2-1=1$ , so $K_p=K_c(RT)^{\Delta n}=4\times10^{-4}\times0.0831\times1000=0.0332\approx0.033$ .

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2025 · NEET-UG (India) · 4 marks

If the molar conductivity ( $\Lambda_m$ ) of a 0.050 $\mathrm{mol\,L^{-1}}$ solution of a monobasic weak acid is 90 $\mathrm{S\,cm^2\,mol^{-1}}$ , its extent (degree) of dissociation will be [Assume $\lambda^\circ_{+} = 349.6\ \mathrm{S\,cm^2\,mol^{-1}}$ and $\lambda^\circ_{-} = 50.4\ \mathrm{S\,cm^2\,mol^{-1}}$ ]
(a) 0.225 (b) 0.215 (c) 0.115 (d) 0.125

Answer

(a) $0.225$

$\Lambda_m^\circ=\lambda^\circ_++\lambda^\circ_-=349.6+50.4=400\ \mathrm{S\,cm^2\,mol^{-1}}$ . $\alpha=\dfrac{\Lambda_m}{\Lambda_m^\circ}=\dfrac{90}{400}=0.225$ .

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2025 · NEET-UG (India) · 4 marks

5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
(a) The solution is ideal. (b) The solution has volume greater than the sum of individual volumes. (c) The solution shows positive deviation. (d) The solution shows negative deviation.

Answer

(d) The solution shows negative deviation.

If ideal, $P=x_XP_X^\circ+x_YP_Y^\circ=\dfrac{5}{15}(63)+\dfrac{10}{15}(78)=21+52=73$ torr. The observed pressure, 70 torr, is lower than 73, so the deviation is negative. This also means the volume is less than additive, which rules out (b).

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2025 · NEET-UG (India) · 4 marks

If the half-life ( $t_{1/2}$ ) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to
(a) 5 minutes (b) 10 minutes (c) 2 minutes (d) 4 minutes

Answer

(b) 10 minutes

$k=\dfrac{0.693}{1\ \mathrm{min}}$ . For 99.9% completion, $t=\dfrac{2.303}{k}\log\dfrac{100}{0.1}=\dfrac{2.303\times3}{0.693}\approx9.97$ min. This is about 10 minutes (roughly 10 half-lives, since $2^{10}\approx1000$ ).

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2025 · NEET-UG (India) · 4 marks

Higher yield of NO in $\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}$ can be obtained at [ $\Delta H$ of the reaction $= +180.7\ \mathrm{kJ\,mol^{-1}}$ ] A. highest temperature B. lower temperature C. higher concentration of $\mathrm{N_2}$ D. higher concentration of $\mathrm{O_2}$ Choose the correct answer from the options given below.
(a) B, C, D only (b) A, C, D only (c) A, D only (d) B, C only

Answer

(b) A, C, D only

The reaction is endothermic ( $\Delta H=+180.7$ kJ), so by Le Chatelier's principle high temperature shifts the equilibrium forward. B (lower temperature) is wrong. Increasing the concentration of either reactant, $\mathrm{N_2}$ or $\mathrm{O_2}$ , also favours more NO. $\Delta n_g=0$ , so pressure has no effect.

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Past questions and answers

3.The correct thermodynamic conditions for the spontaneous reaction at all temperatures is: (a) ΔH<0\Delta H<0ΔH<0 and ΔS=0\Delta S=0ΔS=0 (b) ΔH>0\Delta H>0ΔH>0 and ΔS<0\Delta S<0ΔS<0 (c) ΔH<0\Delta H<0ΔH<0 and ΔS>0\Delta S>0ΔS>0 (d) ΔH<0\Delta H<0ΔH<0 and ΔS<0\Delta S<0ΔS<0

4.Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape? (a) 1/8 (b) 1/4 (c) 3/8 (d) 1/2

6.The pressure of H2\mathrm{H_2}H2​ required to make the potential of H2\mathrm{H_2}H2​-electrode zero in pure water at 298 K is (a) 10−1410^{-14}10−14 atm (b) 10−1210^{-12}10−12 atm (c) 10−1010^{-10}10−10 atm (d) 10−410^{-4}10−4 atm

7.The addition of a catalyst during a chemical reaction alters which of the following quantities? (a) Entropy (b) Internal energy (c) Enthalpy (d) Activation energy

9.At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb=0.52K_b=0.52Kb​=0.52, the boiling point of this solution will be (a) 101 °C (b) 100 °C (c) 102 °C (d) 103 °C

11.During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is (a) 55 minutes (b) 110 minutes (c) 220 minutes (d) 330 minutes

13.The van't Hoff factor (iii) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (a) 0 (b) 1 (c) 2 (d) 3

19.Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because (a) zinc is lighter than iron (b) zinc has lower melting point than iron (c) zinc has lower negative electrode potential than iron (d) zinc has higher negative electrode potential than iron

20.A gas such as carbon monoxide would be most likely to obey the ideal gas law at: (a) low temperatures and low pressures. (b) high temperatures and low pressures. (c) low temperatures and high pressures. (d) high temperatures and high pressures.

21.What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? (a) 1.04 (b) 12.65 (c) 2.0 (d) 7.0

24.The heat of combustion of carbon to CO2\mathrm{CO_2}CO2​ is −393.5-393.5−393.5 kJ/mol. The heat released upon formation of 35.2 g of CO2\mathrm{CO_2}CO2​ from carbon and oxygen gas is: (a) −3.15-3.15−3.15 kJ (b) −315-315−315 kJ (c) +315+315+315 kJ (d) −630-630−630 kJ

25.The rate constant of the reaction A→B\mathrm{A\rightarrow B}A→B is 0.6×10−30.6\times10^{-3}0.6×10−3 mole per second. If the concentration of A is 5 M, then concentration of B after 20 minutes is: (a) 0.72 M (b) 1.08 M (c) 3.60 M (d) 0.36 M

27.Aqueous solution of which of the following compounds is the best conductor of electric current? (a) Fructose, C6H12O6\mathrm{C_6H_{12}O_6}C6​H12​O6​ (b) Acetic acid, C2H4O2\mathrm{C_2H_4O_2}C2​H4​O2​ (c) Hydrochloric acid, HCl (d) Ammonia, NH3\mathrm{NH_3}NH3​

28.What is the mole fraction of the solute in a 1.00 m aqueous solution? (a) 0.0177 (b) 0.177 (c) 1.770 (d) 0.0354

29.Which of the following salts will give highest pH in water? (a) NaCl (b) Na2CO3\mathrm{Na_2CO_3}Na2​CO3​ (c) CuSO4\mathrm{CuSO_4}CuSO4​ (d) KCl

33.The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O2\mathrm{O_2}O2​ at STP will be: (a) 10.8 g (b) 54.0 g (c) 108.0 g (d) 5.4 g

36.When 0.1 mol MnO42−\mathrm{MnO_4^{2-}}MnO42−​ is oxidised the quantity of electricity required to completely oxidise MnO42−\mathrm{MnO_4^{2-}}MnO42−​ to MnO4−\mathrm{MnO_4^-}MnO4−​ is: (a) 2×965002\times965002×96500 C (b) 9650 C (c) 96.50 C (d) 96500 C

37.Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? (a) C6H12O6\mathrm{C_6H_{12}O_6}C6​H12​O6​ (b) Al2(SO4)3\mathrm{Al_2(SO_4)_3}Al2​(SO4​)3​ (c) K2SO4\mathrm{K_2SO_4}K2​SO4​ (d) KCl

39.A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be (a) 0.059 V (b) 0.59 V (c) 0.118 V (d) 1.18 V

40.A reaction having equal energies of activation for forward and reverse reactions has (a) ΔS=0\Delta S=0ΔS=0 (b) ΔG=0\Delta G=0ΔG=0 (c) ΔH=0\Delta H=0ΔH=0 (d) ΔH=ΔG=ΔS=0\Delta H=\Delta G=\Delta S=0ΔH=ΔG=ΔS=0

44.Maximum deviation from ideal gas is expected from (a) H2(g)\mathrm{H_2(g)}H2​(g) (b) N2(g)\mathrm{N_2(g)}N2​(g) (c) CH4(g)\mathrm{CH_4(g)}CH4​(g) (d) NH3(g)\mathrm{NH_3(g)}NH3​(g)

45.Dipole-induced dipole interactions are present in which of the following pairs? (a) H2O\mathrm{H_2O}H2​O and alcohol (b) Cl2\mathrm{Cl_2}Cl2​ and CCl4\mathrm{CCl_4}CCl4​ (c) HCl and He atoms (d) SiF4\mathrm{SiF_4}SiF4​ and He atoms

46.6.02×10206.02\times10^{20}6.02×1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M

48.If molality of the dilute solution is doubled, the value of molal depression constant (KfK_fKf​) will be (a) Doubled (b) Halved (c) Tripled (d) Unchanged

52.Which of the following is dependent on temperature? (a) Molality (b) Molarity (c) Mole fraction (d) Weight percentage

54.A first order reaction has a specific reaction rate of 10−2 s−110^{-2}\ \mathrm{s^{-1}}10−2 s−1. How much time will it take for 20 g of the reactant to reduce to 5 g? (a) 238.6 second (b) 138.6 second (c) 346.5 second (d) 693.0 second

62.When initial concentration of the reactant is doubled, the half-life period of a zero order reaction (a) Is halved (b) Is tripled (c) Is doubled (d) Remains unchanged

65.The correction factor 'a' to the ideal gas equation corresponds to (a) Density of the gas molecules (b) Electric field present between the gas molecules (c) Volume of the gas molecules (d) Forces of attraction between the gas molecules

70.Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (Given that 1 L bar = 100 J) (a) 5 kJ (b) 25 J (c) 30 J (d) −30-30−30 J

71.The mixture that forms maximum boiling azeotrope is: (a) Ethanol + Water (b) Acetone + Carbon disulphide (c) Heptane + Octane (d) Water + Nitric acid

73.If the rate constant for a first order reaction is kkk, the time (ttt) required for the completion of 99% of the reaction is given by: (a) t=6.909/kt = 6.909/kt=6.909/k (b) t=4.606/kt = 4.606/kt=4.606/k (c) t=2.303/kt = 2.303/kt=2.303/k (d) t=0.693/kt = 0.693/kt=0.693/k

76.In which case change in entropy is negative? (a) Expansion of a gas at constant temperature (b) Sublimation of solid to gas (c) 2H(g)→H2(g)2\mathrm{H(g)} \rightarrow \mathrm{H_2(g)}2H(g)→H2​(g) (d) Evaporation of water

80.A mixture of N2\mathrm{N_2}N2​ and Ar gases in a cylinder contains 7 g of N2\mathrm{N_2}N2​ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2\mathrm{N_2}N2​ is: (a) 15 bar (b) 18 bar (c) 9 bar (d) 12 bar

81.The rate constant for a first order reaction is 4.606×10−3 s−14.606\times10^{-3}\ \mathrm{s^{-1}}4.606×10−3 s−1. The time required to reduce 2.0 g of the reactant to 0.2 g is: (a) 500 s (b) 1000 s (c) 100 s (d) 200 s

85.On electrolysis of dil. sulphuric acid using platinum (Pt) electrode, the product obtained at anode will be: (a) H2S\mathrm{H_2S}H2​S gas (b) SO2\mathrm{SO_2}SO2​ gas (c) Hydrogen gas (d) Oxygen gas

86.The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2\mathrm{CaCl_2}CaCl2​ (Atomic mass of Ca = 40 g mol−1\mathrm{mol^{-1}}mol−1) is: (a) 3 (b) 4 (c) 1 (d) 2

88.An increase in the concentration of the reactants of a reaction leads to change in: (a) Threshold energy (b) Collision frequency (c) Activation energy (d) Heat of reaction

89.The mixture which shows positive deviation from Raoult's law is: (a) Acetone + Chloroform (b) Chloroethane + Bromoethane (c) Ethanol + Acetone (d) Benzene + Toluene

92.The pKbpK_bpKb​ of dimethylamine and pKapK_apKa​ of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethyl ammonium acetate solution is: (a) 7.75 (b) 6.25 (c) 8.50 (d) 5.50

94.Which one among the following is the correct option for right relationship between CPC_PCP​ and CVC_VCV​ for one mole of ideal gas? (a) CP=RCVC_P = RC_VCP​=RCV​ (b) CV=RCPC_V = RC_PCV​=RCP​ (c) CP+CV=RC_P + C_V = RCP​+CV​=R (d) CP−CV=RC_P - C_V = RCP​−CV​=R

96.Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O2\mathrm{O_2}O2​ and 2 g H2\mathrm{H_2}H2​ confined in a total volume of one litre at 0∘C0^\circ\mathrm{C}0∘C is: (a) 25.18 (b) 26.02 (c) 2.518 (d) 2.602

102.The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given pKapK_apKa​ of CH3COOH=4.57\mathrm{CH_3COOH} = 4.57CH3​COOH=4.57] (a) 5.57 (b) 3.57 (c) 4.57 (d) 2.57

105.In one molal solution that contains 0.5 mole of a solute, there is (a) 500 mL of solvent (b) 500 g of solvent (c) 100 mL of solvent (d) 1000 g of solvent

107.For a first order reaction A →\rightarrow→ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min−1\mathrm{min^{-1}}min−1 is (a) 1.3818 (b) 0.9212 (c) 0.4606 (d) 0.2303

113.Amongst the given options which of the following molecules/ion acts as a Lewis acid? (a) NH3\mathrm{NH_3}NH3​ (b) H2O\mathrm{H_2O}H2​O (c) BF3\mathrm{BF_3}BF3​ (d) OH−\mathrm{OH^-}OH−

121.Activation energy of any chemical reaction can be calculated if one knows the value of (a) Probability of collision. (b) Orientation of reactant molecules during collision. (c) rate constant at two different temperatures. (d) rate constant at standard temperature.

125.Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is: (Given: Molar mass of Cu: 63 g mol−1\mathrm{mol^{-1}}mol−1, 1 F = 96487 C) (a) 0.315 g (b) 31.5 g (c) 0.0315 g (d) 3.15 g

131.Which of the following aqueous solution will exhibit highest boiling point? (a) 0.01 M Na2SO4\mathrm{Na_2SO_4}Na2​SO4​ (b) 0.015 M C6H12O6\mathrm{C_6H_{12}O_6}C6​H12​O6​ (c) 0.01 M Urea (d) 0.01 M KNO3\mathrm{KNO_3}KNO3​

133.Match List-I (Example) with List-II (Type of Solution). List-I A. Humidity B. Alloys C. Amalgams D. Smoke List-II I. Solid in solid II. Liquid in gas III. Solid in gas IV. Liquid in solid (a) A-III, B-I, C-IV, D-II (b) A-III, B-II, C-I, D-IV (c) A-II, B-IV, C-I, D-III (d) A-II, B-I, C-IV, D-III

137.If the half-life (t1/2t_{1/2}t1/2​) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to (a) 5 minutes (b) 10 minutes (c) 2 minutes (d) 4 minutes

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is: (a) $\Delta H<0$ and $\Delta S=0$ (b) $\Delta H>0$ and $\Delta S<0$ (c) $\Delta H<0$ and $\Delta S>0$ (d) $\Delta H<0$ and $\Delta S<0$

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape? (a) 1/8 (b) 1/4 (c) 3/8 (d) 1/2

The pressure of $\mathrm{H_2}$ required to make the potential of $\mathrm{H_2}$ -electrode zero in pure water at 298 K is (a) $10^{-14}$ atm (b) $10^{-12}$ atm (c) $10^{-10}$ atm (d) $10^{-4}$ atm

The addition of a catalyst during a chemical reaction alters which of the following quantities? (a) Entropy (b) Internal energy (c) Enthalpy (d) Activation energy

At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_b=0.52$ , the boiling point of this solution will be (a) 101 °C (b) 100 °C (c) 102 °C (d) 103 °C

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is (a) 55 minutes (b) 110 minutes (c) 220 minutes (d) 330 minutes

The van't Hoff factor ( $i$ ) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (a) 0 (b) 1 (c) 2 (d) 3

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because (a) zinc is lighter than iron (b) zinc has lower melting point than iron (c) zinc has lower negative electrode potential than iron (d) zinc has higher negative electrode potential than iron

A gas such as carbon monoxide would be most likely to obey the ideal gas law at: (a) low temperatures and low pressures. (b) high temperatures and low pressures. (c) low temperatures and high pressures. (d) high temperatures and high pressures.

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? (a) 1.04 (b) 12.65 (c) 2.0 (d) 7.0

The heat of combustion of carbon to $\mathrm{CO_2}$ is $-393.5$ kJ/mol. The heat released upon formation of 35.2 g of $\mathrm{CO_2}$ from carbon and oxygen gas is: (a) $-3.15$ kJ (b) $-315$ kJ (c) $+315$ kJ (d) $-630$ kJ

The rate constant of the reaction $\mathrm{A\rightarrow B}$ is $0.6\times10^{-3}$ mole per second. If the concentration of A is 5 M, then concentration of B after 20 minutes is: (a) 0.72 M (b) 1.08 M (c) 3.60 M (d) 0.36 M

Aqueous solution of which of the following compounds is the best conductor of electric current? (a) Fructose, $\mathrm{C_6H_{12}O_6}$ (b) Acetic acid, $\mathrm{C_2H_4O_2}$ (c) Hydrochloric acid, HCl (d) Ammonia, $\mathrm{NH_3}$

What is the mole fraction of the solute in a 1.00 m aqueous solution? (a) 0.0177 (b) 0.177 (c) 1.770 (d) 0.0354

Which of the following salts will give highest pH in water? (a) NaCl (b) $\mathrm{Na_2CO_3}$ (c) $\mathrm{CuSO_4}$ (d) KCl

The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of $\mathrm{O_2}$ at STP will be: (a) 10.8 g (b) 54.0 g (c) 108.0 g (d) 5.4 g

When 0.1 mol $\mathrm{MnO_4^{2-}}$ is oxidised the quantity of electricity required to completely oxidise $\mathrm{MnO_4^{2-}}$ to $\mathrm{MnO_4^-}$ is: (a) $2\times96500$ C (b) 9650 C (c) 96.50 C (d) 96500 C

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? (a) $\mathrm{C_6H_{12}O_6}$ (b) $\mathrm{Al_2(SO_4)_3}$ (c) $\mathrm{K_2SO_4}$ (d) KCl

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be (a) 0.059 V (b) 0.59 V (c) 0.118 V (d) 1.18 V

A reaction having equal energies of activation for forward and reverse reactions has (a) $\Delta S=0$ (b) $\Delta G=0$ (c) $\Delta H=0$ (d) $\Delta H=\Delta G=\Delta S=0$

Maximum deviation from ideal gas is expected from (a) $\mathrm{H_2(g)}$ (b) $\mathrm{N_2(g)}$ (c) $\mathrm{CH_4(g)}$ (d) $\mathrm{NH_3(g)}$

Dipole-induced dipole interactions are present in which of the following pairs? (a) $\mathrm{H_2O}$ and alcohol (b) $\mathrm{Cl_2}$ and $\mathrm{CCl_4}$ (c) HCl and He atoms (d) $\mathrm{SiF_4}$ and He atoms

$6.02\times10^{20}$ molecules of urea are present in 100 mL of its solution. The concentration of solution is (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M

If molality of the dilute solution is doubled, the value of molal depression constant ( $K_f$ ) will be
(a) Doubled (b) Halved (c) Tripled (d) Unchanged

Which of the following is dependent on temperature?
(a) Molality (b) Molarity (c) Mole fraction (d) Weight percentage

A first order reaction has a specific reaction rate of $10^{-2}\ \mathrm{s^{-1}}$ . How much time will it take for 20 g of the reactant to reduce to 5 g?
(a) 238.6 second (b) 138.6 second (c) 346.5 second (d) 693.0 second

When initial concentration of the reactant is doubled, the half-life period of a zero order reaction
(a) Is halved (b) Is tripled (c) Is doubled (d) Remains unchanged

The correction factor 'a' to the ideal gas equation corresponds to
(a) Density of the gas molecules (b) Electric field present between the gas molecules (c) Volume of the gas molecules (d) Forces of attraction between the gas molecules

Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (Given that 1 L bar = 100 J)
(a) 5 kJ (b) 25 J (c) 30 J (d) $-30$ J

The mixture that forms maximum boiling azeotrope is:
(a) Ethanol + Water (b) Acetone + Carbon disulphide (c) Heptane + Octane (d) Water + Nitric acid

If the rate constant for a first order reaction is $k$ , the time ( $t$ ) required for the completion of 99% of the reaction is given by:
(a) $t = 6.909/k$ (b) $t = 4.606/k$ (c) $t = 2.303/k$ (d) $t = 0.693/k$

In which case change in entropy is negative?
(a) Expansion of a gas at constant temperature (b) Sublimation of solid to gas (c) $2\mathrm{H(g)} \rightarrow \mathrm{H_2(g)}$ (d) Evaporation of water

A mixture of $\mathrm{N_2}$ and Ar gases in a cylinder contains 7 g of $\mathrm{N_2}$ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of $\mathrm{N_2}$ is:
(a) 15 bar (b) 18 bar (c) 9 bar (d) 12 bar

The rate constant for a first order reaction is $4.606\times10^{-3}\ \mathrm{s^{-1}}$ . The time required to reduce 2.0 g of the reactant to 0.2 g is:
(a) 500 s (b) 1000 s (c) 100 s (d) 200 s

On electrolysis of dil. sulphuric acid using platinum (Pt) electrode, the product obtained at anode will be:
(a) $\mathrm{H_2S}$ gas (b) $\mathrm{SO_2}$ gas (c) Hydrogen gas (d) Oxygen gas

The number of Faradays (F) required to produce 20 g of calcium from molten $\mathrm{CaCl_2}$ (Atomic mass of Ca = 40 g $\mathrm{mol^{-1}}$ ) is:
(a) 3 (b) 4 (c) 1 (d) 2

An increase in the concentration of the reactants of a reaction leads to change in:
(a) Threshold energy (b) Collision frequency (c) Activation energy (d) Heat of reaction

The mixture which shows positive deviation from Raoult's law is:
(a) Acetone + Chloroform (b) Chloroethane + Bromoethane (c) Ethanol + Acetone (d) Benzene + Toluene

The $pK_b$ of dimethylamine and $pK_a$ of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethyl ammonium acetate solution is:
(a) 7.75 (b) 6.25 (c) 8.50 (d) 5.50

Which one among the following is the correct option for right relationship between $C_P$ and $C_V$ for one mole of ideal gas?
(a) $C_P = RC_V$ (b) $C_V = RC_P$ (c) $C_P + C_V = R$ (d) $C_P - C_V = R$

Choose the correct option for the total pressure (in atm.) in a mixture of 4 g $\mathrm{O_2}$ and 2 g $\mathrm{H_2}$ confined in a total volume of one litre at $0^\circ\mathrm{C}$ is:
(a) 25.18 (b) 26.02 (c) 2.518 (d) 2.602

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given $pK_a$ of $\mathrm{CH_3COOH} = 4.57$ ]
(a) 5.57 (b) 3.57 (c) 4.57 (d) 2.57

In one molal solution that contains 0.5 mole of a solute, there is
(a) 500 mL of solvent (b) 500 g of solvent (c) 100 mL of solvent (d) 1000 g of solvent

For a first order reaction A $\rightarrow$ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in $\mathrm{min^{-1}}$ is
(a) 1.3818 (b) 0.9212 (c) 0.4606 (d) 0.2303

Amongst the given options which of the following molecules/ion acts as a Lewis acid?
(a) $\mathrm{NH_3}$ (b) $\mathrm{H_2O}$ (c) $\mathrm{BF_3}$ (d) $\mathrm{OH^-}$

Activation energy of any chemical reaction can be calculated if one knows the value of
(a) Probability of collision. (b) Orientation of reactant molecules during collision. (c) rate constant at two different temperatures. (d) rate constant at standard temperature.

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is: (Given: Molar mass of Cu: 63 g $\mathrm{mol^{-1}}$ , 1 F = 96487 C)
(a) 0.315 g (b) 31.5 g (c) 0.0315 g (d) 3.15 g

Which of the following aqueous solution will exhibit highest boiling point?
(a) 0.01 M $\mathrm{Na_2SO_4}$ (b) 0.015 M $\mathrm{C_6H_{12}O_6}$ (c) 0.01 M Urea (d) 0.01 M $\mathrm{KNO_3}$

Match List-I (Example) with List-II (Type of Solution). List-I A. Humidity B. Alloys C. Amalgams D. Smoke List-II I. Solid in solid II. Liquid in gas III. Solid in gas IV. Liquid in solid
(a) A-III, B-I, C-IV, D-II (b) A-III, B-II, C-I, D-IV (c) A-II, B-IV, C-I, D-III (d) A-II, B-I, C-IV, D-III

If the half-life ( $t_{1/2}$ ) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to
(a) 5 minutes (b) 10 minutes (c) 2 minutes (d) 4 minutes