The vapour pressure of acetone at 20$$^\circ$$C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20$$^\circ$$C, its vapour pressure was 183 torr. The molar mass (g mol$$^{-1}$$) of the substance is:

The pure solvent is acetone. At $$20^{\circ}\text C$$ its vapour pressure is given as $$P^{\circ}=185\ \text{torr}$$.

After dissolving a non-volatile solute the observed vapour pressure of the solution becomes $$P_{\text{soln}}=183\ \text{torr}$$. Because the solute is non-volatile, only acetone contributes to the vapour of the solution, so we can apply Raoult’s law for a dilute solution.

Raoult’s law for a non-volatile solute states first:

$$\frac{P^{\circ}-P_{\text{soln}}}{P^{\circ}}=x_{\text{solute}}$$

where $$x_{\text{solute}}$$ is the mole fraction of the solute in the liquid phase.

We start by evaluating the left-hand side, called the relative lowering of vapour pressure:

$$P^{\circ}-P_{\text{soln}}=185-183=2\ \text{torr}$$

Hence

$$\frac{P^{\circ}-P_{\text{soln}}}{P^{\circ}}=\frac{2}{185}=0.01081$$

Therefore

$$x_{\text{solute}}=0.01081=\frac{2}{185}$$

Next we must express the mole fraction in terms of the unknown molar mass of the solute. Let that molar mass be $$M\ (\text{g mol}^{-1}).$$

Mass of solute given: $$1.2\ \text{g}$$, so

$$n_{\text{solute}}=\frac{1.2}{M}\ \text{mol}$$

Mass of acetone taken: $$100\ \text{g}$$. Its molar mass is $$58\ \text{g mol}^{-1}$$ (because $$ \text{CH}_3\text{CO}\text{CH}_3 $$ has $$3\times12+6\times1+16 = 58$$). Thus

$$n_{\text{acetone}}=\frac{100}{58}=1.7241\ \text{mol}$$

The mole fraction of the solute is, by definition,

$$x_{\text{solute}}=\frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{acetone}}} =\frac{\dfrac{1.2}{M}}{\dfrac{1.2}{M}+\dfrac{100}{58}}$$

We equate this expression to the value obtained from Raoult’s law:

$$\frac{\dfrac{1.2}{M}}{\dfrac{1.2}{M}+\dfrac{100}{58}}=\frac{2}{185}$$

To clear the complex fraction we denote

$$S=\frac{1.2}{M},\quad L=\frac{100}{58}=1.7241$$

This converts the equation into

$$\frac{S}{S+L}=\frac{2}{185}$$

Cross-multiplying gives

$$185\,S=2\,(S+L)$$

$$185\,S=2S+2L$$

Subtracting $$2S$$ from both sides:

$$183\,S=2L$$

So

$$S=\frac{2L}{183}$$

Substituting back $$S=\dfrac{1.2}{M}$$ and $$L=1.7241$$, we have

$$\frac{1.2}{M}=\frac{2\times1.7241}{183}$$

Evaluating the right-hand side:

$$2\times1.7241=3.4482$$

$$\frac{3.4482}{183}=0.0188446$$

Thus

$$\frac{1.2}{M}=0.0188446$$

Hence

$$M=\frac{1.2}{0.0188446}=63.7\ \text{g mol}^{-1}$$

The value rounds to $$64\ \text{g mol}^{-1}$$, which matches option C.

Hence, the correct answer is Option C.