0.1 molal aqueous solution of glucose boils at $100.16^{\circ} \mathrm{C}$. What is boiling point of 0.5…
- $100.80^{\circ} \mathrm{C}$
- $100.16^{\circ} \mathrm{C}$
- $100.10^{\circ} \mathrm{C}$
- $20.8^{\circ} \mathrm{C}$
Solution
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.
Asked in: MHT CET 2024 (10 May Shift 1)