Question

Two solvent A and B have $K_b$ values 1.8 and 2.7 K mol$^{-1}$ kg respectively. A given amount of substance when dissolved in 500 g of A, completely dimerized and when same amount of the substance is dissolved in 500 g of B the solute shows trimerization. If ratio of observed elevation in boiling point in A and B is 1: 2. Find the percentage extent of trimerization.

• A. 30%
• B. 40%
• C. 50%
• D. 60%

Answer 1

Answer: (C)

50%

Answer 2

The problem involves the colligative property of elevation in boiling point ($\Delta T_b$) and the concept of van't Hoff factor ($i$) for association.

1. Formula for Elevation in Boiling Point: The elevation in boiling point is given by: $\Delta T_b = i \cdot K_b \cdot m$ where:

• $i$ is the van't Hoff factor.
• $K_b$ is the ebullioscopic constant of the solvent.
• $m$ is the molality of the solution.

2. Molality (m): Let the given amount of substance correspond to $n$ moles. The mass of solvent A = 500 g = 0.5 kg. The mass of solvent B = 500 g = 0.5 kg. Since the amount of substance and the mass of solvent are the same for both solutions, the molality ($m$) will be the same for both. $m = \frac{n \text{ moles}}{0.5 \text{ kg}}$

3. Van't Hoff Factor ($i$) for Solvent A (Dimerization): In solvent A, the substance completely dimerizes. The reaction is $2X \rightleftharpoons (X)_2$. If we start with 1 mole of X, upon complete dimerization, we get $1/2$ mole of $(X)_2$. The van't Hoff factor $i_A = \frac{\text{moles of particles after association}}{\text{moles of particles before association}} = \frac{1/2}{1} = 0.5$. Given $K_b(A) = 1.8$ K mol$^{-1}$ kg. So, $\Delta T_b(A) = i_A \cdot K_b(A) \cdot m = 0.5 \cdot 1.8 \cdot m = 0.9m$.

4. Van't Hoff Factor ($i$) for Solvent B (Trimerization): In solvent B, the solute shows trimerization. The reaction is $3X \rightleftharpoons (X)_3$. Let $\alpha$ be the extent of trimerization. If we start with 1 mole of X: Initial moles: $X = 1$, $(X)_3 = 0$. Moles at equilibrium: $X = 1 - \alpha$, $(X)_3 = \alpha/3$. Total moles at equilibrium = $(1 - \alpha) + \alpha/3 = 1 - \frac{3\alpha - \alpha}{3} = 1 - \frac{2\alpha}{3}$. The van't Hoff factor $i_B = \frac{\text{total moles at equilibrium}}{\text{initial moles}} = \frac{1 - 2\alpha/3}{1} = 1 - 2\alpha/3$. Given $K_b(B) = 2.7$ K mol$^{-1}$ kg. So, $\Delta T_b(B) = i_B \cdot K_b(B) \cdot m = (1 - 2\alpha/3) \cdot 2.7 \cdot m$.

5. Using the Ratio of Elevation in Boiling Points: The ratio of observed elevation in boiling point in A and B is given as 1:2. $\frac{\Delta T_b(A)}{\Delta T_b(B)} = \frac{1}{2}$ Substitute the expressions for $\Delta T_b(A)$ and $\Delta T_b(B)$: $\frac{0.9m}{(1 - 2\alpha/3) \cdot 2.7m} = \frac{1}{2}$ The molality 'm' cancels out: $\frac{0.9}{(1 - 2\alpha/3) \cdot 2.7} = \frac{1}{2}$ Cross-multiply: $0.9 \cdot 2 = (1 - 2\alpha/3) \cdot 2.7$ $1.8 = (1 - 2\alpha/3) \cdot 2.7$ Divide both sides by 2.7: $1 - 2\alpha/3 = \frac{1.8}{2.7} = \frac{18}{27} = \frac{2}{3}$ Now, solve for $\alpha$: $2\alpha/3 = 1 - 2/3$ $2\alpha/3 = 1/3$ Multiply by 3: $2\alpha = 1$ $\alpha = 0.5$

6. Percentage Extent of Trimerization: The extent of trimerization ($\alpha$) is 0.5. Percentage extent of trimerization = $\alpha \times 100\% = 0.5 \times 100\% = 50\%$.