Q37 / 150 · MHT CET 2024 (10 May Shift 1) · Chemistry
0.1 molal aqueous solution of glucose boils at $100.16^{\circ} \mathrm{C}$. What is boiling point of 0.5 molal aqueous solution of glucose?
- Correct answerA$100.80^{\circ} \mathrm{C}$
- B$100.16^{\circ} \mathrm{C}$
- C$100.10^{\circ} \mathrm{C}$
- D$20.8^{\circ} \mathrm{C}$
Solution
Given:
- 0.1 molal solution boils at $100.16^{\circ} \mathrm{C}$.
- Boiling point elevation formula:
$\Delta T_b=i K_b \cdot m$
where $m$ is molality.
Step 1: Elevation for 0.1 molal solution:
$\Delta T_b=100.16-100=0.16^{\circ} C$
Using $\Delta T_b \propto m$, for 0.5 molal solution: $\Delta T_b^{\prime}=0.16 \cdot \frac{0.5}{0.1}=0.8^{\circ} \mathrm{C}$
Step 2: New Boiling Point: $T_b^{\prime}=100+0.8=100.80^{\circ} \mathrm{C}$
Answer: $100.80^{\circ} C$, Option 1.
Using $\Delta T_b \propto m$, for 0.5 molal solution: $\Delta T_b^{\prime}=0.16 \cdot \frac{0.5}{0.1}=0.8^{\circ} \mathrm{C}$
Step 2: New Boiling Point: $T_b^{\prime}=100+0.8=100.80^{\circ} \mathrm{C}$
Answer: $100.80^{\circ} C$, Option 1.
Hence, the correct answer is option: 'A'.