1. Use Raoult’s law for a non-volatile solute

For a solution containing a non-volatile solute,

$P = X_{\text{solvent}} P^0$

So,

$\frac{P^0 - P}{P^0} = X_{\text{solute}}$

Given:

• Vapour pressure of pure acetone, $P^0 = 185\ \text{torr}$
• Vapour pressure of solution, $P = 183\ \text{torr}$

Hence relative lowering of vapour pressure is

$\frac{185-183}{185} = \frac{2}{185}$

Thus,

$X_{\text{solute}} = \frac{2}{185}$

2. Write mole fraction in terms of moles

Let moles of solute be $n_2$ and moles of acetone be $n_1$. Then,

$X_{\text{solute}} = \frac{n_2}{n_1+n_2}$

Since the solute amount is small, we may also use the exact relation directly.

3. Calculate moles of acetone

Molar mass of acetone $= 58\ \text{g mol}^{-1}$

Mass of acetone $= 100\ \text{g}$

$n_1 = \frac{100}{58} = 1.724\ \text{mol}$

4. Find moles of solute exactly

Using

$\frac{n_2}{n_1+n_2} = \frac{2}{185}$

Substitute $n_1 = \frac{100}{58}$:

$\frac{n_2}{\frac{100}{58}+n_2} = \frac{2}{185}$

Cross-multiplying,

$185n_2 = 2\left(\frac{100}{58}+n_2\right)$

$185n_2 = \frac{200}{58} + 2n_2$

$183n_2 = \frac{200}{58}$

$n_2 = \frac{200}{58\times 183}$

$n_2 \approx 0.01884\ \text{mol}$

5. Calculate molar mass of solute

Mass of solute $= 1.2\ \text{g}$

$M_2 = \frac{1.2}{0.01884} \approx 63.7\ \text{g mol}^{-1}$

So the molar mass is approximately

$64\ \text{g mol}^{-1}$

6. Check options

• A: 64 ✅
• B: 128 ❌
• C: 488 ❌
• D: 32 ❌

Therefore, the correct option is A.