Question

\(0.05\) mole of a non-volatile solute is dissolved in \(500\,g\) of water. What is the depression in freezing point of resultant solution?
\((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)

• A. \(0.047\,K\)
• B. \(0.372\,K\)
• C. \(0.093\,K\)
• D. \(0.186\,K\)

Hint:

For freezing point depression: \[ \Delta T_f=K_fm \] where molality is \[ m=\frac{\text{moles of solute}}{\text{kg of solvent}} \] Always convert solvent mass into kilograms before calculation.

Answer 1

The Correct Option is D

Step 1: Recall the formula for depression in freezing point.
The depression in freezing point is given by $\Delta T_f = K_f \times m$, where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.
Step 2: Identify the given data.
Moles of solute $= 0.05$ mol. Mass of solvent (water) $= 500$ g $= 0.5$ kg. $K_f = 1.86$ K$\cdot$kg/mol. The solute is non-volatile and we assume it is non-electrolyte (van't Hoff factor $i = 1$).
Step 3: Calculate the molality.
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05}{0.5} = 0.1 \text{ mol/kg} \]
Step 4: Apply the freezing point depression formula.
\[ \Delta T_f = K_f \times m = 1.86 \times 0.1 = 0.186 \text{ K} \]
Step 5: Match with the options.
Checking the options: 0.047 K (no), 0.372 K (no), 0.093 K (no), 0.186 K (yes). Option 4 matches our calculated value.
Step 6: Confirm the answer.
The depression in freezing point is $0.186$ K, which corresponds to option 4. This makes physical sense because dilute solutions (low molality) show small but measurable colligative effects.
\[ \boxed{\Delta T_f = 0.186 \text{ K}} \]