Solutions and Colligative Properties - Result Question 38

####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:

  1. 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,

$$ \begin{array}{rlrl} & \frac{185-183}{183} & =\frac{\frac{1.2}{M}}{\frac{100}{58}} \Rightarrow \frac{2}{183}=\frac{1.2 \times 58}{100 \times M} \ \therefore \quad M & =\frac{183 \times 1.2 \times 58}{2 \times 100} \ M & =63.684=64 g / mol \end{array} $$



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