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Standard XII

Chemistry

Abnormal Colligative Properties

When a liquid...

Question

When a liquid that is immiscible with water was steam distilled at $95 . 2^{o} C$ at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at $95 . 2^{o} C$ , What is the molar mass of liquid?

7.975 g/mol

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166 g/mol

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145.8 g/mol

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None of these

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Solution

The correct option is C 145.8 g/mol
For two immiscible liquid ;
$P_{A}^{\circ} P_{t o t a l}$ $P_{H_{2} O}^{\circ} = 748 - 648 \Rightarrow 100$
$\frac{W_{A}}{W_{B}} = \frac{P_{A}^{\circ} M_{A}}{P_{B}^{\circ} M_{B}}$ : $M_{A} = \frac{1.25}{1} \times \frac{648 \times 18}{100}$

$\Rightarrow 145.8$

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Similar questions

Q. When a liquid that is immiscible with water was steam distilled at

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Q. An unknown liquid 'A' is immiscible with water. It is steam distilled at

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Q. A mixture of an organic liquid

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and water distilled under one atmospheric pressure at

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) was dissolved in 1000 g of liquid

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(molar mass 180 g

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Calculate the vapour pressure of pure liquid

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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

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of liquid B (molar mass

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When a liquid that is immiscible with water was steam distilled at 95.2oC at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2oC, What is the molar mass of liquid?

The correct option is C 145.8 g/molFor two immiscible liquid ;P∘APtotal P∘H2O=748−648⇒100WAWB=P∘AMAP∘BMB : MA=1.251×648×18100⇒145.8

When a liquid that is immiscible with water was steam distilled at $95 . 2^{o} C$ at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at $95 . 2^{o} C$ , What is the molar mass of liquid?

The correct option is C 145.8 g/mol
For two immiscible liquid ;
$P_{A}^{\circ} P_{t o t a l}$ $P_{H_{2} O}^{\circ} = 748 - 648 \Rightarrow 100$
$\frac{W_{A}}{W_{B}} = \frac{P_{A}^{\circ} M_{A}}{P_{B}^{\circ} M_{B}}$ : $M_{A} = \frac{1.25}{1} \times \frac{648 \times 18}{100}$

$\Rightarrow 145.8$