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Multiple Choice Questions

1. Under the same reaction conditions, initial concentration of 1.386 mol of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio of the rate constants for first order and zero order of the reactions is: [JEE 2008 P-I]

(a) $0.5 \text{ mol}^{-1} \text{dm}^3$

(b) $1.0 \text{ mol}^{-1} \text{dm}^3$

(d) $2.0 \text{ mol}^{-1} \text{dm}^3$

Answer: (a)

Explanation:

$$k_1 = \frac{0.693}{t_{1/2}} = \frac{0.693}{40}$$

$$k_0 = \frac{[A]_0}{2t_{1/2}} = \frac{1.386}{2 \times 20}$$

$$\frac{k_1}{k_0} = \frac{0.693 / 40}{1.386 / 40} = \frac{0.693}{1.386} = 0.5 \text{ mol}^{-1} \text{dm}^3$$

2. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is: [JEE 2008 P-II]

(a) $\text{CH}_3(\text{CH}_2)_{15}\text{N}^+(\text{CH}_3)_3\text{Br}^-$

(b) $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$

(d) $\text{CH}_3(\text{CH}_2)_{11}\text{N}^+(\text{CH}_3)_3\text{Br}^-$

Answer: (a)

Explanation: Critical concentration for micelle formation decreases as the molecular weight of hydrocarbon chain of surfactant grows because in this case true solubility diminishes and the tendency of surfactant molecule to associate increases.

3. The correct statement (s) pertaining to the adsorption of a gas on a solid surface is (are) [JEE 2011 P-I]

(a) Adsorption is always exothermic

(b) Physisorption may transform into chemisorption at high temperature

(d) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation.

Answer: (a, b, d)

Explanation: Adsorption is always exothermic ($\Delta H < 0$). Physisorption decreases with increase in temperature, whereas chemisorption increases initially and then decreases with increase in temperature due to activation energy requirement.

4. Choose the correct reason(s) for the stability of the lyophobic colloidal particles. [JEE 2012 P-I]

(a) Preferential adsorption of ions on their surface from the solution

(b) Preferential adsorption of solvent on their surface from the solution

(d) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles

Answer: (a, d)

Explanation:

(a) Preferential adsorption of ions on surface from the solution leads to charge development on colloidal particles.

(d) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles (Zeta Potential) provides electrostatic repulsion between particles preventing coagulation.

5. The given graphs/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct? [JEE 2012 P-II]

(I) Amount of gas adsorbed vs T (at constant P): decreases continuously with T

(II) Amount of gas adsorbed vs T (at constant P): increases to a maximum then decreases with T

(III) $\frac{x}{m} = kP^{1/n}$ (Freundlich adsorption isotherm)

(IV) $\Delta H_{\text{ads}} = 150 \text{ kJ mol}^{-1}$

(a) I is physisorption and II is chemisorptions

(b) I is physisorption and III is chemisorptions

(d) IV is chemisorption and III is chemisorption

Answer: (a, c)

Explanation: I: In physisorption, as the temperature increases the amount of gas adsorbed decreases. II: In chemisorption, as the temperature increases the amount of gas adsorbed increases (to a certain range of temperature) then decreases. III: Freundlich adsorption isotherms for physisorption (at constant T). IV: $\Delta H_{\text{ads}} = 150 \text{ kJ mol}^{-1}$ is in the range for chemisorption ($80-240 \text{ kJ mol}^{-1}$).

6. The coagulating power of electrolytes having ions $\text{Na}^+$, $\text{Al}^{3+}$ and $\text{Ba}^{2+}$ for arsenic sulphide sol increases in the order [JEE Main 2013]

(a) $\text{Al}^{3+} < \text{Ba}^{2+} < \text{Na}^+$

(b) $\text{Na}^+ < \text{Ba}^{2+} < \text{Al}^{3+}$

(d) $\text{Al}^{3+} < \text{Na}^+ < \text{Ba}^{2+}$

Answer: (b)

Explanation: Arsenic sulphide ($\text{As}_2\text{S}_3$) sol is negatively charged. According to Hardy-Schulze rule, higher the valency of the coagulating ion, greater is its coagulating power. Therefore, coagulating power increases as $\text{Na}^+ < \text{Ba}^{2+} < \text{Al}^{3+}$.

7. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is: [JEE Main 2015]

(a) 18 mg

(b) 36 mg

(d) 54 mg

Answer: (a)

Explanation: Amount of acetic acid adsorbed per gram of charcoal:

$$\text{Amount} = \frac{(0.06 - 0.042) \times 50 \times 10^{-3} \times 60}{3} = \frac{0.018 \times 50 \times 10^{-3} \times 60}{3} = 18 \times 10^{-3}\text{ g} = 18\text{ mg}$$

8. When $\text{O}_2$ is adsorbed on a metallic surface, electron transfer occurs from the metal to $\text{O}_2$. The TRUE statement(s) regarding this adsorption is (are): [JEE Adv. 2015 P-II]

(a) $\text{O}_2$ is physisorbed

(b) heat is released

(d) bond length of $\text{O}_2$ is increased

Answer: (b, c, d)

Explanation: Adsorption of $\text{O}_2$ on metal surface involves electron transfer (chemisorption) and is exothermic ($\Delta H < 0$). During electron transfer from metal to $\text{O}_2$, electrons occupy $\pi^* 2p$ antibonding orbital of $\text{O}_2$. Due to electron transfer to antibonding orbital, the bond order of $\text{O}_2$ decreases, hence bond length increases.

9. For a linear plot of $\log (x/m)$ versus $\log p$ in a Freundlich adsorption isotherm, which of the following statements is correct? [JEE Main 2016]

(a) Both k and 1/n appear in the slope term

(b) 1/n appears as the intercept

(d) $\log (1/n)$ appears as the intercept.

Answer: (c)

Explanation: According to Freundlich adsorption isotherm:

$$\frac{x}{m} = k P^{1/n}$$

$$\log \left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log P$$

Comparing with $y = mx + c$, slope $= 1/n$ and intercept $= \log k$. Thus, only $1/n$ appears as the slope.

10. The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of $\text{KCl}$, $\text{CH}_3\text{OH}$ and $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$ at room temperature. The correct assignment of the sketches is [JEE Adv. 2016 P-II]

(a) I : $\text{KCl}$, II : $\text{CH}_3\text{OH}$, III : $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$

(b) I : $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$, II : $\text{CH}_3\text{OH}$, III : $\text{KCl}$

(d) I : $\text{CH}_3\text{OH}$, II : $\text{KCl}$, III : $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$

Answer: (d)

Explanation: Impurities affect surface tension appreciably. Detergents and soaps like $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$ decrease the surface tension sharply (Graph III). Alcohols like $\text{CH}_3\text{OH}$ lower the surface tension slightly (Graph I). Inorganic electrolytes like $\text{KCl}$ increase the surface tension of water (Graph II).

11. The Tyndall effect is observed only when following conditions are satisfied [JEE Main 2017]

A. The diameter of the dispersed particles is much smaller than the wavelength of the light used.

B. The diameter of the dispersed particle is not much smaller than the wavelength of the light used.

C. The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.

D. The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

(a) B and C

(b) A and D

(d) A and C

Answer: (c)

Explanation: Tyndall effect is observed only when: (1) The diameter of the dispersed particles is not much smaller than the wavelength of the light used (Condition B). (2) The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude (Condition D).

12. The correct statements about surface properties is (are): [JEE Adv. 2017 P-II]

(a) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system

(b) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium

(c) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature

(d) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution

Answer: (a, c)

Explanation: (a) Adsorption is an exothermic process ($\Delta H_{\text{sys}} < 0$) and is accompanied by decrease in entropy ($\Delta S_{\text{sys}} < 0$). (c) Higher the critical temperature ($T_c$), stronger are the intermolecular forces of attraction and hence higher is the extent of adsorption. Cloud is an aerosol (liquid in gas), not an emulsion. Brownian motion depends on the size of colloidal particles as well as viscosity.

13. For coagulation of arsenious sulphide sol, which one of the following salt solution will be most effective? [JEE Main 2019]

(a) $\text{AlCl}_3$

(b) $\text{NaCl}$

(d) $\text{Na}_3\text{PO}_4$

Answer: (a)

Explanation: Arsenious sulphide ($\text{As}_2\text{S}_3$) sol is a negatively charged colloid. Thus, the cation with maximum charge will be most effective for coagulation. Coagulating power order: $\text{Al}^{3+} > \text{Ba}^{2+} > \text{Na}^+$.

14. If dichloromethane (DCM) and water ($\text{H}_2\text{O}$) are used for differential extraction, which one of the following statements is correct? [JEE Main 2019]

(a) DCM and $\text{H}_2\text{O}$ would stay as lower and upper layer respectively in the S.F.

(b) DCM and $\text{H}_2\text{O}$ will be miscible clearly

(d) DCM and $\text{H}_2\text{O}$ will make turbid/colloidal mixture

Answer: (a)

Explanation: Dichloromethane ($\text{CH}_2\text{Cl}_2$) is immiscible with water and has a higher density ($\approx 1.33\text{ g/cm}^3$) than water ($\approx 1.0\text{ g/cm}^3$). Therefore, DCM forms the lower layer and $\text{H}_2\text{O}$ forms the upper layer in a separating funnel.

15. The freezing point of a diluted milk sample is found to be $-0.2^\circ\text{C}$, while it should have been $-0.5^\circ\text{C}$ for pure milk. How much water has been added to pure milk to make the diluted sample? [JEE Main 2019]

(a) 2 cups of water to 3 cups of pure milk

(b) 1 cup of water to 3 cups of pure milk

(d) 1 cup of water to 2 cups of pure milk

Answer: (c)

Explanation: $\Delta T_f = K_f \times m \propto \frac{\text{moles of solute}}{\text{mass/volume of water}}$.

$$\frac{\Delta T_f(\text{diluted})}{\Delta T_f(\text{pure})} = \frac{0.2}{0.5} = \frac{V_{\text{pure}}}{V_{\text{diluted}}} \implies \frac{V_{\text{pure}}}{V_{\text{pure}} + V_{\text{added water}}} = \frac{2}{5}$$

$$\implies 5 V_{\text{pure}} = 2 V_{\text{pure}} + 2 V_{\text{added water}} \implies 3 V_{\text{pure}} = 2 V_{\text{added water}} \implies \frac{V_{\text{added water}}}{V_{\text{pure}}} = \frac{3}{2}$$

Thus, 3 cups of water were added to 2 cups of pure milk.

16. Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is: [JEE Main 2019]

(a) C : solid in liquid; M : solid in liquid; S : solid in gas

(b) C : solid in liquid; M : liquid in liquid; S : gas in solid

(d) C : liquid in solid; M : liquid in liquid; S : solid in gas

Answer: (d)

Explanation:

Colloid	Dispersed Phase	Dispersion Medium
Cheese (C)	Liquid	Solid
Milk (M)	Liquid	Liquid
Smoke (S)	Solid	Gas

17. The molecule that has minimum/no role in the formation of photochemical smog is: [JEE Main 2019]

(a) $\text{CH}_2 = \text{O}$

(b) $\text{N}_2$

(d) $\text{NO}$

Answer: (b)

Explanation: Photochemical smog consists of oxides of nitrogen ($\text{NO}$, $\text{NO}_2$), ozone ($\text{O}_3$), formaldehyde ($\text{CH}_2=\text{O}$), acrolein, and PAN. Unreactive atmospheric $\text{N}_2$ gas has minimum/no direct role in its formation.

18. The compound that is not a common component of photochemical smog is: [JEE Main 2019]

(a) $\text{O}_3$

(b) $\text{CH}_2 = \text{CHCHO}$

(d) $\text{H}_3\text{C}-\text{C}(=\text{O})\text{OONO}_2$

Answer: (c)

Explanation: Freons (CFCs like $\text{CF}_2\text{Cl}_2$) are responsible for ozone layer depletion in the stratosphere, but are not common components of photochemical smog.

19. Among the following, the false statement is: [JEE Main 2019]

(a) Latex is a colloidal solution of rubber particles which are positively charged

(b) Tyndall effect can be used to distinguish between a colloidal solution and a true solution.

(c) It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to the one on clouds from an aeroplane.

(d) Lyophilic sol can be coagulated by adding an electrolyte.

Answer: (a)

Explanation: Latex is a colloidal solution of rubber particles which are negatively charged, not positively charged. Statement (a) is false.

Statement/Assertion and Reason

20. Statement-I: Micelles are formed by surfactant molecules above the critical micellar concentration (CMC). And Statement-II: The conductivity of a solution having surfactant molecules decreases sharply at the CMC. [JEE 2007 P-I]

(a) Statement-I is True, Statement-II is True; Statement -II is a correct explanation for Statement-I.

(b) Statement-I is True, Statement-II is True; Statement -II is not a correct explanation for Statement-I.

(d) Statement -I is False, Statement-II is True.

Answer: (b)

Explanation: The formation of micelles takes place only above a particular temperature called Kraft temperature and above a particular concentration called critical micelle concentration (CMC). Above CMC, individual ions aggregate to form single micellar particles, causing the conductivity of the solution to decrease sharply per unit concentration. Both statements are true, but Statement-II is not the correct explanation for Statement-I.

Integer and Subjective

21. 20% surface sites have adsorbed $\text{N}_2$. On heating $\text{N}_2$ gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume $2.46\text{ cm}^3$. Density of surface sites is $6.023 \times 10^{14}/\text{cm}^2$ and surface area is $1000\text{ cm}^2$, find out the no. of surface sites occupied per molecule of $\text{N}_2$. [JEE 2005]

Answer: 2

Explanation:

By ideal gas equation, $PV = nRT$

$$n(\text{N}_2) = \frac{PV}{RT} = \frac{0.001 \times 2.46 \times 10^{-3}}{0.0821 \times 298} = 1.0 \times 10^{-7}\text{ mol}$$

$$\text{Molecules of N}_2 = 6.023 \times 10^{23} \times 1.0 \times 10^{-7} = 6.023 \times 10^{16}$$

Total surface sites available $= 6.023 \times 10^{14} \times 1000 = 6.023 \times 10^{17}$

Surface sites used to adsorb $\text{N}_2 = \frac{20}{100} \times 6.023 \times 10^{17} = 12.04 \times 10^{16}$

$$\text{Sites occupied per molecule} = \frac{12.04 \times 10^{16}}{6.023 \times 10^{16}} = 2$$

22. Silver (atomic weight $= 108\text{ g mol}^{-1}$) has a density of $10.5\text{ g cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12}\text{ m}^2$ can be expressed in scientific notation as $Y \times 10^{-x}$. The value of x is: [JEE 2010 P-II]

Answer: 7

Explanation:

$$\text{Density} = \frac{\text{mass}}{V} = 10.5\text{ g/cm}^3 \implies 1\text{ cm}^3 \text{ has } 10.5\text{ g of Ag}$$

$$\text{Number of atoms of Ag in } 1\text{ cm}^3 = \frac{10.5}{108} N_A$$

$$\text{In } 1\text{ cm}, \text{ number of atoms of Ag} = \left(\frac{10.5}{108} N_A\right)^{1/3}$$

$$\text{In } 1\text{ cm}^2, \text{ number of atoms of Ag} = \left(\frac{10.5}{108} N_A\right)^{2/3}$$

$$\text{In } 10^{-12}\text{ m}^2 = 10^{-8}\text{ cm}^2, \text{ number of atoms of Ag} = \left(\frac{10.5}{108} N_A\right)^{2/3} \times 10^{-8}$$

$$= \left(\frac{10.5 \times 6.022 \times 10^{23}}{108}\right)^{2/3} \times 10^{-8} = 1.5 \times 10^{-7}$$

Comparing with $Y \times 10^{-x}$, we get $x = 7$.