Solutions and Colligative Properties - Result Question 8
####9. 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
$(2015,1 M)$
(a) 32
(b) 64
(a) 128
(b) 488
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Solution:
- Given, $p^{\circ}=185$ Torr at $20^{\circ} C$
$$ p _s=183 \text { Torr at } 20^{\circ} C $$
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,
$$ \begin{array}{rlrl} & \frac{185-183}{183} & =\frac{\frac{1.2}{M}}{\frac{100}{58}} \Rightarrow \frac{2}{183}=\frac{1.2 \times 58}{100 \times M} \ \therefore \quad M & =\frac{183 \times 1.2 \times 58}{2 \times 100} \ M & =63.684=64 g / mol \end{array} $$
