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Standard XIII

Chemistry

Relative Lowering of Vapour Pressure

The vapour pr...

Question

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 $(g m o l^{- 1})$ of the substance is:

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128

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488

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Solution

The correct option is B
64

$\frac{P^{\circ} - P}{P^{\circ}} = X_{s o l u t e} = \frac{n_{1}}{n_{1} + n_{2}} \frac{185 - 183}{185} = \frac{2}{185} = \frac{\frac{1.2}{M}}{\frac{1.2}{M} + \frac{100}{58}} \Rightarrow M = 64$

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Similar questions

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 $(g m o l^{- 1})$ of the substance is:

Q. The vapour Pressure of acetone at

20^{\circ} C

is 185 torr. When 1.2 g 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

Q. 100 g of liquid

A

(molar mass 140 g

m o l^{- 1}

) was dissolved in 1000 g of liquid

B

(molar mass 180 g

m o l^{- 1}

) The vapour pressure of pure liquid

B

was found to be 500 torr.

Calculate the vapour pressure of pure liquid

A

and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr:

100 g

of liquid A (molar mass

140 g m o l^{- 1})

was dissolved in

1000 g

of liquid B (molar mass

180 g m o l^{- 1})

. The vapour pressure of pure liquid B was found to be

500 t o r r

. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is

475 t o r r

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The vapour pressure of acetone at 20∘C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20∘C, its vapour pressure was 183 torr. The molar mass (gmol−1) of the substance is:

The correct option is B 64 P∘−PP∘=Xsolute=n1n1+n2185−183185=2185=1.2M1.2M+10058⇒M=64

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 $(g m o l^{- 1})$ of the substance is:

The correct option is B
64

$\frac{P^{\circ} - P}{P^{\circ}} = X_{s o l u t e} = \frac{n_{1}}{n_{1} + n_{2}} \frac{185 - 183}{185} = \frac{2}{185} = \frac{\frac{1.2}{M}}{\frac{1.2}{M} + \frac{100}{58}} \Rightarrow M = 64$