Vapour pressure of $C_{6} H_{6}$ at $80^{\circ} C$ in lowered by dissolving 2gm of a non–volatile substance in 78 gm of Benzene. The vapour pressure of pure $C_{6} H_{6}$ at $80^{\circ} C$ is 750 mm. The molecular lot. Of the substance is

100

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150

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200

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250

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Solution

The correct option is B 150
$\frac{p^{\circ} - p}{p^{\circ}} = \frac{w}{m} \times \frac{M}{W} \frac{10}{750} = \frac{2}{m} \times \frac{78}{78}$
m = 150

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Vapour pressure of C6H6 at 80∘C in lowered by dissolving 2gm of a non–volatile substance in 78 gm of Benzene. The vapour pressure of pure C6H6 at 80∘C is 750 mm. The molecular lot. Of the substance is

The correct option is B 150p∘−pp∘=wm×MW10750=2m×7878 m = 150

Vapour pressure of $C_{6} H_{6}$ at $80^{\circ} C$ in lowered by dissolving 2gm of a non–volatile substance in 78 gm of Benzene. The vapour pressure of pure $C_{6} H_{6}$ at $80^{\circ} C$ is 750 mm. The molecular lot. Of the substance is

The correct option is B 150
$\frac{p^{\circ} - p}{p^{\circ}} = \frac{w}{m} \times \frac{M}{W} \frac{10}{750} = \frac{2}{m} \times \frac{78}{78}$
m = 150