Question

Calculate the molality of the solution of nonvolatile solute if it freezes at $-0.36^{\circ}\text{C}$. [$K_{f}$ for solvent $=1.86\text{ K kg mol}^{-1}$]}

• A. 0.218 $\text{mol kg}^{-1}$
• B. 0.193 $\text{mol kg}^{-1}$
• C. 0.401 $\text{mol kg}^{-1}$
• D. 0.520 $\text{mol kg}^{-1}$

Hint:

Depression in freezing point ($\Delta T_f$) is always a positive magnitude! Just drop the negative sign from the solution's freezing point when plugging it into the equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the freezing point of an aqueous solution containing a non-volatile solute and the cryoscopic constant ($K_f$) of the solvent. We need to calculate the molality ($m$) of the solution.

Step 2: Detailed Explanation:
The normal freezing point of the pure solvent (water) is $0^{\circ}\text{C}$.
The freezing point of the solution is given as $T_f = -0.36^{\circ}\text{C}$.
The depression in freezing point ($\Delta T_f$) is defined as:
$\Delta T_f = T_{f(\text{pure})} - T_{f(\text{solution})}$
$\Delta T_f = 0^{\circ}\text{C} - (-0.36^{\circ}\text{C}) = 0.36^{\circ}\text{C}$ (or $0.36 \text{ K}$)
The colligative property formula relating freezing point depression to molality is:
$\Delta T_f = K_f \times m$
(Assuming the van 't Hoff factor $i = 1$ because it states a "nonvolatile solute" without mentioning dissociation).
Rearrange to solve for molality ($m$):
$m = \frac{\Delta T_f}{K_f}$
Substitute the known values:
$m = \frac{0.36}{1.86}$
Perform the division:
$m \approx 0.1935 \text{ mol kg}^{-1}$
Rounding to three decimal places matches $0.193 \text{ mol kg}^{-1}$.

Step 3: Final Answer:
The molality is 0.193 $\text{mol kg}^{-1}$, matching option (b).