Question

The vapour pressure of acetone at $20^{\circ }$C is $185$ torr .When $1.2g$ 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

Answer: (B)

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