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

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Solution

The correct option is B 64
$p^{0} = 185 T o r r a t 20^{\circ} C$
$p_{s} = 183 T o r r a t 20^{\circ} C$
Mass of non- volatile substance, m = 1.2 g
Mass of acetone taken = 100g
M = ?
As, we have $\frac{p^{0} - p_{s}}{p_{s}} = \frac{n}{N}$
Putting the values, we get,
$\frac{185 - 183}{183} \frac{\frac{1.2}{M}}{\frac{100}{58}} \Rightarrow \frac{2}{183} = \frac{1.2 \times 58}{100 \times M}$
$∴ M = \frac{183.1.2 \times 58}{2 \times 100}$
$M = 63.684 = 64 g / m o l$

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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

100

g of acetone at

20^{\circ}

C, its vapour pressure was

183

torr.

The molar mass (

g m o l

^{- 1}

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100 g of liquid A (molar mass 140 g mol⁻¹) was dissolved in 1000 g of liquid B (molar mass 180 g mol⁻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.

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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 100g of acetone at 20∘C, its vapour pressure was 183 Torr. The molar mass of the substance is

The correct option is B 64p0=185 Torr at 20∘C ps=183 Torr at 20∘C Mass of non- volatile substance, m = 1.2 g Mass of acetone taken = 100g M = ? As, we have p0−psps=nN Putting the values, we get, 185−1831831.2M10058⇒2183=1.2×58100×M ∴ M=183.1.2×582×100 M=63.684=64g/mol

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