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 100g of acetone at ${20}^{\circ }C$, its vapour pressure was 183 Torr. The molar mass of the substance is

• A. 32
• B. 64
• C. 128
• D. 488

Answer 1

The correct option is B 64

$p^0=185Torrat{20}^{\circ }C$
$p_s=183Torrat{20}^{\circ }C$
Mass of non- volatile substance, m = 1.2 g
Mass of acetone taken = 100g
M = ?
As, we have $\frac{p^0-p_s}{p_s}=\frac{n}{N}$
Putting the values, we get,
$\frac{185-183}{183}\frac{\frac{1.2}{M}}{\frac{100}{58}}\Rightarrow \frac{2}{183}=\frac{1.2\times 58}{100\times M}$
$\therefore M=\frac{\mathrm{183.1.2}\times 58}{2\times 100}$
$M=63.684=64g/mol$