Question

0.1 molal aqueous solution of glucose boils at 100.16°C. What is boiling point of 0.5 molal aqueous solution of glucose?

• A. 500.80°C
• B. 100.80°C
• C. 20.16°C
• D. 20.8°C

Answer 1

Answer: (B)

100.80°C

Answer 2

We use the boiling point elevation formula:

$$\Delta T_b = K_b \times m$$

1. For the 0.1 molal solution:

   • Normal boiling point of water = 100°C
   • Observed boiling point = 100.16°C
   • Elevation, $\Delta T_b = 100.16 - 100 = 0.16^\circ\text{C}$

2. Calculate the molal boiling point elevation constant:

   $$K_b = \frac{0.16^\circ\text{C}}{0.1\,m} = 1.6^\circ\text{C/m}$$

3. For the 0.5 molal solution:

   $$\Delta T_b = 1.6^\circ\text{C/m} \times 0.5\,m = 0.80^\circ\text{C}$$

4. Therefore, the boiling point of the 0.5 molal solution is:

   $$100^\circ\text{C} + 0.80^\circ\text{C} = 100.80^\circ\text{C}$$