Bihar Board Class 12 Science Chemistry · Ch 1: Solutions

Topics in this test

Difficulty: 3 easy · 8 medium · 19 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. At 300 K, two separate aqueous solutions are prepared: Solution X is $0.1\,m$ glucose and Solution Y is $0.1\,m$ $\text{NaCl}$ (assume complete dissociation). If $K_f$ for water is $1.86\ \text{K kg mol}^{-1}$, which statement about their freezing-point depression is correct?

   • A. Both have equal $\Delta T_f$ as molality is same
   • B. $\Delta T_f$ of Y is half that of X
   • C. $\Delta T_f$ of Y is twice that of X
   • D. $\Delta T_f$ of X is twice that of Y
   Solution

   For NaCl the van't Hoff factor $i=2$, so $\Delta T_f = i\,K_f\,m = 2 \times 1.86 \times 0.1 = 0.372$ K, while glucose (non-electrolyte, $i=1$) gives $0.186$ K — exactly half. More dissolved particles means a greater depression.

2. Q2. Consider the following statements about an ideal solution: (I) It obeys Raoult's law over the entire range of concentration. (II) The enthalpy of mixing ($\Delta H_{mix}$) is zero. (III) The volume of mixing ($\Delta V_{mix}$) is zero. Which statements are correct?

   • A. Only II and III
   • B. I, II and III
   • C. Only I and II
   • D. Only I
   Solution

   For an ideal solution, solute–solvent interactions equal the pure component interactions, so it obeys Raoult's law with $\Delta H_{mix} = 0$ and $\Delta V_{mix} = 0$.

3. Q3. Assertion (A): The experimentally determined molar mass of acetic acid dissolved in benzene is nearly double its normal value. Reason (R): Acetic acid molecules undergo association (dimerisation) in benzene. Choose the correct option.

   • A. A is false but R is true
   • B. A is true but R is false
   • C. Both A and R are true but R is not the correct explanation of A
   • D. Both A and R are true and R is the correct explanation of A
   Solution

   Two acetic acid molecules join through hydrogen bonds to form a dimer, reducing the number of particles and giving a higher (abnormal) molar mass; R correctly explains A.

4. Q4. The vapour pressure of pure water at a certain temperature is $23.8\,mm\,Hg$. If the mole fraction of a non-volatile solute in an aqueous solution is $0.2$, the vapour pressure of the solution will be:

   • A. $23.80\,mm\,Hg$
   • B. $4.76\,mm\,Hg$
   • C. $28.56\,mm\,Hg$
   • D. $19.04\,mm\,Hg$
   Solution

   By Raoult's law, $p = p^\circ \times x_{solvent} = 23.8 \times (1 - 0.2) = 23.8 \times 0.8 = 19.04\,mm\,Hg$.

5. Q5. When $1\,g$ of a non-volatile, non-electrolyte solute is dissolved in $50\,g$ of water, the freezing point is lowered by $0.186\,K$. If $K_f$ for water is $1.86\,K\,kg\,mol^{-1}$, the molar mass of the solute is:

   • A. $200\,g\,mol^{-1}$
   • B. $100\,g\,mol^{-1}$
   • C. $180\,g\,mol^{-1}$
   • D. $60\,g\,mol^{-1}$
   Solution

   Using $M = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} = \frac{1.86 \times 1 \times 1000}{0.186 \times 50} = 200\,g\,mol^{-1}$.

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