Question

6 g of a non-volatile solute (x) is dissolved in 100 g of water. The relative lowering of vapour pressure of resultant solution is 0.006. What is the molar mass (in g mol\textsuperscript{–1}) of x?

• A. 60
• B. 360
• C. 100
• D. 180

Hint:

Use the formula for relative lowering of vapour pressure: \(\frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}\) for molar mass calculations.

Answer 1

The Correct Option is D

Relative lowering of vapour pressure = \(\frac{P^0 - P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}\)
Given:
- Mass of solute = 6 g
- Mass of water = 100 g ⇒ moles = \(\frac{100}{18}\)
- Lowering of vapour pressure = 0.006
Let molar mass be \(M\). Then, \[ \frac{6/M}{100/18} = 0.006 \Rightarrow \frac{6 \times 18}{100 \times 0.006} = M \Rightarrow M = 180 \text{ g mol}^{-1} \]