Question:
An aqueous solution of a non-volatile and non-electrolytic solute boils at $100.5^{\circ}C.$ What will be the freezing point of the same solution? (Given $K_{b}=0.512~K$ kg $mol^{-1}$ and $K_{f}=1.86~K$ kg mol$^{-1}$)
An aqueous solution of a non-volatile and non-electrolytic solute boils at $100.5^{\circ}C.$ What will be the freezing point of the same solution? (Given $K_{b}=0.512~K$ kg $mol^{-1}$ and $K_{f}=1.86~K$ kg mol$^{-1}$)
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Taking the ratio $\frac{\Delta T_{f}}{\Delta T_{b}} = \frac{K_{f}}{K_{b}}$ saves time by bypassing the calculation of molality entirely.
Updated On: Jun 3, 2026
- $-2.816^{\circ}C$
- $-1.816^{\circ}C$
- $-0.908^{\circ}C$
- $-3.632^{\circ}C$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The elevation of boiling point ($\Delta T_{b}$) and depression of freezing point ($\Delta T_{f}$) are colligative properties that depend on the molality ($m$) of the solution.
Step 2: Meaning
Since the boiling point of pure water is $100^{\circ}C$, the elevation in boiling point is: $$\Delta T_{b} = 100.5^{\circ}C - 100^{\circ}C = 0.5^{\circ}C$$
Step 3: Analysis
The formulas for the two colligative properties are: $$\Delta T_{b} = K_{b} \cdot m$$ $$\Delta T_{f} = K_{f} \cdot m$$ Dividing the two equations eliminates molality ($m$): $$\frac{\Delta T_{f}}{\Delta T_{b}} = \frac{K_{f}}{K_{b}}$$ Substituting the given values: $$\frac{\Delta T_{f}}{0.5} = \frac{1.86}{0.512}$$ $$\Delta T_{f} = \frac{1.86 \cdot 0.5}{0.512} = \frac{0.93}{0.512} \approx 1.816^{\circ}C$$ The freezing point of the solution ($T_{f}$) is calculated from the freezing point of pure water ($0^{\circ}C$): $$T_{f} = 0^{\circ}C - \Delta T_{f} = 0^{\circ}C - 1.816^{\circ}C = -1.816^{\circ}C$$
Step 4: Conclusion
Thus, the freezing point of the solution is $-1.816^{\circ}C$.
Final Answer: (B)
The elevation of boiling point ($\Delta T_{b}$) and depression of freezing point ($\Delta T_{f}$) are colligative properties that depend on the molality ($m$) of the solution.
Step 2: Meaning
Since the boiling point of pure water is $100^{\circ}C$, the elevation in boiling point is: $$\Delta T_{b} = 100.5^{\circ}C - 100^{\circ}C = 0.5^{\circ}C$$
Step 3: Analysis
The formulas for the two colligative properties are: $$\Delta T_{b} = K_{b} \cdot m$$ $$\Delta T_{f} = K_{f} \cdot m$$ Dividing the two equations eliminates molality ($m$): $$\frac{\Delta T_{f}}{\Delta T_{b}} = \frac{K_{f}}{K_{b}}$$ Substituting the given values: $$\frac{\Delta T_{f}}{0.5} = \frac{1.86}{0.512}$$ $$\Delta T_{f} = \frac{1.86 \cdot 0.5}{0.512} = \frac{0.93}{0.512} \approx 1.816^{\circ}C$$ The freezing point of the solution ($T_{f}$) is calculated from the freezing point of pure water ($0^{\circ}C$): $$T_{f} = 0^{\circ}C - \Delta T_{f} = 0^{\circ}C - 1.816^{\circ}C = -1.816^{\circ}C$$
Step 4: Conclusion
Thus, the freezing point of the solution is $-1.816^{\circ}C$.
Final Answer: (B)
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