Question

5 g sucrose (molar mass = 342) is dissolved in 100 g of solvent, decreases the freezing point by 2.15 K. What is cryoscopic constant of solvent?

• A. $14.7\ \text{K}\ \text{kg}\ \text{mol}^{-1}$
• B. $2.15\ \text{K}\ \text{kg}\ \text{mol}^{-1}$
• C. $4.30\ \text{K}\ \text{kg}\ \text{mol}^{-1}$
• D. $7.35\ \text{K}\ \text{kg}\ \text{mol}^{-1}$

Hint:

To perform this calculation quickly without long division, notice that $\frac{2.15}{5} = 0.43$. Then simply compute $0.43 \times 34.2 \approx 14.7$ mentally.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the mass and molar mass of a solute (sucrose), the mass of the solvent, and the resulting depression in freezing point ($\Delta T_f$). We need to determine the cryoscopic constant ($K_f$) of the solvent.

Step 2: Key Formula or Approach:
The depression in freezing point is a colligative property governed by the formula: $$\Delta T_f = K_f \times m$$ where $m$ represents the molality of the solution. Expanding molality gives the full expression: $$\Delta T_f = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1}$$ Rearranging this equation to isolate the cryoscopic constant ($K_f$): $$K_f = \frac{\Delta T_f \times M_2 \times W_1}{1000 \times W_2}$$

Step 3: Detailed Explanation:
Let's list the values provided: Mass of solute (sucrose), $W_2 = 5\ \text{g}$ Molar mass of solute, $M_2 = 342\ \text{g}\ \text{mol}^{-1}$ Mass of solvent, $W_1 = 100\ \text{g}$ Freezing point depression, $\Delta T_f = 2.15\ \text{K}$ Substitute these values into our rearranged equation: $$K_f = \frac{2.15 \times 342 \times 100}{1000 \times 5}$$ Simplify the terms in the fraction: $$K_f = \frac{2.15 \times 342}{10 \times 5} = \frac{735.3}{50}$$ $$K_f = \frac{14.706}{1} \approx 14.7\ \text{K}\ \text{kg}\ \text{mol}^{-1}$$

Step 4: Final Answer:
The cryoscopic constant of the solvent is $14.7\ \text{K}\ \text{kg}\ \text{mol}^{-1}$, which matches option (A).