Question

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 ($gmol$${}^{-1}$) of the substance is:

• A. $32$
• B. $64$
• C. $128$
• D. $488$

Answer 1

The correct option is B $64$

The molar mass $M_2$ of non-volatile substance can be calculated from the following formula.

$\frac{p^0-p}{p^0}=\frac{W_2\times M_1}{W_1\times M_2}$

Here, $p^0$ vapour pressure of acetone and $p$ is the vapour pressure of the solution.

$M_1$ and $W_1$ are the molar mass and mass of acetone.

$M_2$ and $W_2$ are the molar mass and mass of a non-volatile substance.

Substitute values in the above expression.

$\frac{185-183}{185}=\frac{1.2\times 58}{100\times M_2}$

Hence, $M_2=64.38g/mol$.

Thus, the molar mass of the non-volatile substance is $64$ $g/mol$ (approximately).

Hence, the correct option is $\text{B}$