Question

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer) 
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$] 
Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Hint:

For dilute aqueous solutions, you can use the relation $RLVP = \frac{m \cdot M_{solvent}}{1000}$. This allows you to bypass calculating individual mole counts.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
Elevation in boiling point ($\Delta T_{b}$) and relative lowering of vapour pressure (RLVP) are both colligative properties.

Step 2: Key Formula or Approach:
1. $\Delta T_{b} = K_{b} \cdot m$.
2. $RLVP = \frac{P^{\circ} - P}{P^{\circ}} = X_{solute}$.
3. For dilute solutions, $X_{solute} \approx \frac{n_{solute}}{n_{solvent}}$.

Step 3: Detailed Explanation:
Find molality ($m$) from boiling point elevation:
\[ 0.5 = 5.0 \times m \implies m = 0.1 \text{ mol kg}^{-1} \]
Molality is moles of solute per kg of solvent:
\[ n_{solute} = m \times W_{solvent (kg)} = 0.1 \times 0.150 = 0.015 \text{ mol} \]
Now, find moles of solvent ($n_{solvent}$):
\[ n_{solvent} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{150}{300} = 0.5 \text{ mol} \]
Calculate RLVP:
\[ RLVP = \frac{n_{solute}}{n_{solute} + n_{solvent}} \approx \frac{0.015}{0.5} = 0.03 \]
\[ RLVP = 3 \times 10^{-2} \]

Step 4: Final Answer:
The relative lowering in vapour pressure is $3 \times 10^{-2}$. The integer is 3.