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Standard XI

Chemistry

Relative Lowering of Vapour Pressure

The vapour pr...

Question

The vapour pressure of acetone at $20^{\circ}$ $C$ is 185 $t o r r$ . 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 $t o r r$ . The molar mass $(g / m o l)$ 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
We know that,
$\frac{P_{o} - P_{s}}{P_{s}} = \frac{n}{N}$
$\frac{185 - 183}{183}$ = $\frac{\frac{1.2}{M}}{\frac{100}{58}}$

$M \approx 64 g / m o l$

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Similar questions

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

QUESTION 2.36

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

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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 (g/mol) of the substance is:

The correct option is B 64We know that, Po−PsPs=nN 185−183183 = 1.2M10058 M≈64g/mol

The vapour pressure of acetone at $20^{\circ}$ $C$ is 185 $t o r r$ . 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 $t o r r$ . The molar mass $(g / m o l)$ of the substance is:

The correct option is B 64
We know that,
$\frac{P_{o} - P_{s}}{P_{s}} = \frac{n}{N}$
$\frac{185 - 183}{183}$ = $\frac{\frac{1.2}{M}}{\frac{100}{58}}$

$M \approx 64 g / m o l$