Question

At 50 °C, the vapour pressure of pure benzene is 268 torr. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of 167 torr at the same temperature is

• A. 0.505
• B. 0.705
• C. 0.605
• D. 0.405

Answer 1

The Correct Option is C

To solve the problem, we use Raoult’s Law which relates the vapour pressure of a solution to the mole fraction of the solvent.

1. Given Data:
- Vapour pressure of pure benzene ($P^0$) = 268 torr
- Vapour pressure of solution ($P_{solution}$) = 167 torr
- Moles of benzene = 1 mol (as per question)
- Molar mass of benzene = 78 g/mol (not directly needed for this calculation)

2. Apply Raoult’s Law:
Raoult’s Law: $P_{solution} = X_{benzene} \cdot P^0$
Where $X_{benzene}$ is the mole fraction of benzene:
$X_{benzene} = \frac{n_{benzene}}{n_{benzene} + n_{solute}}$
Let moles of solute = $x$, and moles of benzene = 1 mol
\[ 167 = \left(\frac{1}{1 + x}\right) \cdot 268 \]

3. Solve for x:

\[ \frac{1}{1 + x} = \frac{167}{268} = 0.6231 \] \[ 1 + x = \frac{1}{0.6231} \approx 1.6048 \Rightarrow x = 1.6048 - 1 = 0.6048 \approx 0.605 \]

Final Answer:
The number of moles of non-volatile solute required per mole of benzene is 0.605.