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 of the substance is

• A. 32
• B. 64
• C. 128
• D. 488

Answer 1

Answer: (B)

Given, $p^{\circ}=185$ Torr at $20^{\circ} C$
$p_{s}=183$ Torr at $20^{\circ} C$
Mass of non-volatile substance, $m=1.2 \,g$
Mass of acetone taken $=100 \,g$
$M=?$
As, we have $\frac{p^{\circ}-p_{s}}{p_{s}}=\frac{n}{N}$
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{183 \times 1.2 \times 58}{2 \times 100} $
$M =63.684=64\, g / mol$