Surface Chemistry JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Surface Chemistry, free to read — no sign-in needed. The full chapter has 16 questions; sign in to attempt the remaining 11 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from $10 \mathrm{mg} \mathrm{g}^{-1}$ and $16 \mathrm{mg} \mathrm{g}^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be $4 \mathrm{mg} \mathrm{g}^{-1}$ and $10 \mathrm{mg} \mathrm{g}^{-1}$, respectively. At this temperature, the concentration (in $\mathrm{mg} \mathrm{g}^{-1}$ ) of adsorbed phenol from $20 \mathrm{mg} \mathrm{g}^{-1}$ aqueous solution of phenol will be $\qquad$ . Use : $\log _{10} 2=0.3$
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   Answer: 15.62

   $\begin{aligned} & \frac{x}{m}=K \times C^{1 / n} \\ & \log \left(\frac{x}{m}\right)=\log K+\frac{1}{n} \log C \\ & \log 4=\log K+\frac{1}{n} \log 10 \\ & 0.6=\log K+\frac{1}{n} .....(1) \\ & \log 10=\log K+\frac{1}{n} \log 16 \\ & 1=\log K+\frac{1}{n} \times 1.2 ....(2) \end{aligned}$ Equation (2) - equation (1) $0.4=\frac{1}{\mathrm{n}} \times(0.2) \Rightarrow \mathrm{n}=0.5$ and $\log \mathrm{K}=-1.4$ $\begin{aligned} & \log \frac{x}{m}=\log K+\frac{1}{n} \times \log C \\ & =-1.4+2 \times \log 20 \end{aligned}$ $\begin{aligned} & =-1.4+2.6=1.2 \\ & \frac{\mathrm{x}}{\mathrm{m}}=10^{+1.2}=16 \\ & \left(\log 2=0.3,4 \log 2=1.2,16=10^{+1.2}\right)\end{aligned}$ $\begin{aligned} & \frac{x}{m}=K \times C^{1 / n} \\ & 4=K(10)^{1 / n}....(1) \\ & 10=K(16)^{1 / n}....(2) \\ & X=K(20)^{1 / n}....(3) \end{aligned}$ On solving equation (1) and (2) $\frac{1}{n}=2$ On solving equation (1) and (3) $\frac{4}{X}=\left(\frac{10}{20}\right)^2$ $X=16$ On solving equation (2) and (3) $\begin{aligned} & \frac{10}{X}=\left(\frac{16}{20}\right)^2 \\ & X=15.625 \end{aligned}$
2. Q2JEE Advanced Adv 2023 (Paper 2)
   Consider the following statements related to colloids. (I) Lyophobic colloids are not formed by simple mixing of dispersed phase and dispersion medium. (II) For emulsions, both the dispersed phase and the dispersion medium are liquid. (III) Micelles are produced by dissolving a surfactant in any solvent at any temperature. (IV) Tyndall effect can be observed from a colloidal solution with dispersed phase having the same refractive index as that of the dispersion medium. The option with the correct set of statements is
   1. A.(I) and (II)
   2. B.(II) and (III)
   3. C.(III) and (IV)
   4. D.(II) and (IV)
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   Answer: (A)

   As in Lyophobic colloids there is no interaction between dispersed phase and dispersion medium, special methods are used for preparation, simple mixing will not form colloid. Emulsion are colloids of liquid dispersed phase and liquid dispersion medium. Micelles formation occurs when temperature is above a particular temperature called Kraft temperature ${T}_{K}$ and concentration above a particular value know as critical micelle concentration (CMC). Tyndall's effect can be observed when the refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
3. Q3JEE Advanced Adv 2022 (Paper 1)
   The correct option(s) related to adsorption processes is(are)
   1. A.Chemisorption results in a unimolecular layer.
   2. B.The enthalpy change during physisorption is in the range of $100$ to $140kJ{mol}^{-1}$.
   3. C.Chemisorption is an endothermic process.
   4. D.Lowering the temperature favours physisorption processes.
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   Answer: A,D

   (A) First statement is correct as chemisorption results in a unimolecular layer and physisorption result in a multimolecular layer. (B) Second statement is incorrect as enthalpy change during physisorption is of the range of $\left(20-40\right){kJmol}^{-1}$.Gases with higher critical temperatures are readily adsorbed (C) Chemisorption is an exothermic process with $\left(80-240\right){kJmol}^{-1}$ as the enthalpy of adsorption. (D) Lowering the temperature results in increase in the extent of physisorption.It decreases with increase of temperature. Hence (A) and (D) are correct.
4. Q4JEE Advanced Adv 2021 (Paper 1)
   The correct statement $\left(s\right)$ related to colloids is(are)
   1. A.The process of precipitating colloidal sol by an electrolyte is called peptization.
   2. B.Colloidal solution freezes at higher temperature than the true solution at the same concentration.
   3. C.Surfactants form micelle above critical micelle concentration $\left(CMC\right).CMC$ depends on temperature.
   4. D.Micelles are macromolecular colloids.
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   Answer: B,C

   The process of setting of colloidal particles by adding electrolyte is called coagulation or precipitation of the sol.Peptization is the process of converting a precipitate into colloidal sol by shaking it in a medium in presence of a small amount of electrolyte. Colloidal solution have a lower value of colligative properly so, their depression in freezing point will be low, and freezing point will be higher. Surfactants form micelle above C.M.C and value of CMC depends on temperature. Micelles are not macromolecular colloids they are associated colloids. Since, many small molecules associate with each other to form micelles.
5. Q5JEE Advanced Adv 2017 (Paper 2)
   The correct statement(s) about surface properties is(are)
   1. A.Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium
   2. B.Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system
   3. C.Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution
   4. D.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
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   Answer: B,D

   (i) Emulsion is liquid in liquid type colloid. (ii) For adsorption,$\Delta H\lt 0$ and $\Delta S\lt 0$ (iii) Smaller the size and less viscous the dispersion medium, more will be the Brownian motion. (iv) Higher the ${T}_{C}$, greater will be the extent of adsorption.