Solutions and Colligative Properties 1 Question 8
9. The vapour pressure of acetone at $20^{\circ} \mathrm{C}$ is 185 torr.
When $1.2 \mathrm{~g}$ of a non-volatile substance was dissolved in $100 \mathrm{~g}$ of acetone at $20^{\circ} \mathrm{C}$, its vapour pressure was 183 Torr. The molar mass of the substance is
$(2015,1 \mathrm{M})$
(a) 32
(b) 64
(a) 128
(b) 488
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Solution:
- Given, $p^{\circ}=185$ Torr at $20^{\circ} \mathrm{C}$
$$ p_{s}=183 \text { Torr at } 20^{\circ} \mathrm{C} $$
Mass of non-volatile substance, $m=1.2 \mathrm{~g}$
Mass of acetone taken $=100 \mathrm{~g}$
$$ M=? $$
As, we have $\frac{p^{\circ}-p_{s}}{p_{s}}=\frac{n}{N}$
Putting the values, we get,
$$ \begin{array}{rlrl} & \frac{185-183}{183} & =\frac{\frac{1.2}{M}}{\frac{100}{58}} \Rightarrow \frac{2}{183}=\frac{1.2 \times 58}{100 \times M} \ \therefore \quad M & =\frac{183 \times 1.2 \times 58}{2 \times 100} \ M & =63.684=64 \mathrm{~g} / \mathrm{mol} \end{array} $$
