Question

An aqueous solution containing 3 g of a solute of molar mass 111.6 g mol\textsuperscript{--1 in a certain mass of water freezes at --0.125 $^\circ$C. The mass of water in grams present in the solution is ($K_f = 1.86$ K kg mol\textsuperscript{--1})}

• A. 300
• B. 600
• C. 500
• D. 400
• E. 250

Hint:

Always ensure $\Delta T_f$ is a positive value. It is the difference between the freezing point of the pure solvent and the solution.

Answer 1

The Correct Option is D

Concept: Depression in freezing point ($\Delta T_f$) is a colligative property that depends on the molality of the solution.
• Formula: $\Delta T_f = K_f \times m$, where $m$ is molality.
• Molality ($m$): $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{w_B \times 1000}{M_B \times w_A \text{ (in grams)}}$.

Step 1: Identify the known variables. - $\Delta T_f = 0 - (-0.125) = 0.125 \text{ K}$ - $w_B$ (mass of solute) = 3 g - $M_B$ (molar mass of solute) = 111.6 g/mol - $K_f = 1.86 \text{ K kg mol}^{-1}$

Step 2: Solve for the mass of water ($w_A$). Substitute variables into the combined formula: \[ \Delta T_f = K_f \times \frac{w_B \times 1000}{M_B \times w_A} \] \[ 0.125 = 1.86 \times \frac{3 \times 1000}{111.6 \times w_A} \] \[ w_A = \frac{1.86 \times 3000}{111.6 \times 0.125} = \frac{5580}{13.95} \] \[ w_A = 400 \text{ g} \]