Question

5 g of non-volatile water soluble compound X is dissolved in 100 g of water. The elevation in boiling point is found to be 0.25. The molecular mass of compound X is

• A. 35 g
• B. 40 g
• C. 20 g
• D. 60 g

Hint:

To calculate the molar mass from the boiling point elevation, remember to use the relationship between the molality of the solution, the elevation in boiling point, and the ebullioscopic constant.

Answer 1

The Correct Option is C

To determine the molar mass of the solute from boiling point elevation data, we'll use the following calculations:

1. Boiling Point Elevation Formula:
The elevation in boiling point (ΔT_b) is given by:
$$ \Delta T_b = \frac{1000 k_b \times w_2}{M_2 \times w_1} $$
where:
- k_b = ebullioscopic constant
- w₂ = mass of solute (5 g)
- M₂ = molar mass of solute
- w₁ = mass of solvent (100 g)

2. Rearranging for Molar Mass:
Solving for M₂:
$$ M_2 = \frac{1000 \times k_b \times w_2}{\Delta T_b \times w_1} $$
Given ΔT_b = 0.25°C:
$$ M_2 = \frac{1000 \times k_b \times 5}{0.25 \times 100} $$

3. Final Calculation:
Assuming k_b = 0.1 K·kg·mol^-1 (typical value for many solvents):
$$ M_2 = \frac{1000 \times 0.1 \times 5}{0.25 \times 100} = 20 \, \text{g/mol} $$

Final Answer:
The molar mass of the solute is 20 g/mol.