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Standard XII

Chemistry

Raoult's Law

A solution is...

Question

A solution is prepared by mixing 8.5 g of $C H_{2} C l_{2}$ and 11.95 g of $C H C l_{3}$ . If vapour pressure of $C H_{2} C l_{2}$ and $C H C l_{3}$ at 298 K are 415 and 200 mmHg respectively, the mole fraction of $C H C l_{3}$ in vapour form is:

[Molar mass of C1=35.5 g $m o l^{- 1}$ ]

0.325

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0.486

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0.675

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0.162

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Solution

The correct option is A 0.325
Given

$P_{0} =$ Vapor pressure = 200 mm of Hg

Mass of solute (CH $_{2}$ Cl $_{2}$ )=8.5 g

Mass of solvent (CHCl $_{3}$ )=11.95 g

Mole fraction in solution phase can be calculated as:

$X_{C H_{2} C l_{2}} = \frac{(8.5 / 85)}{(8.5 / 85) + (11.95 / 119.5)} = 0.5$

$X_{C H C l_{3}} = \frac{(11.95 / 119.5)}{(8.5 / 85) + (11.95 / 119.5)} = 0.5$

Mole fraction in vapor phase is given by the formula:

$Y_{C H C l_{3}} = \frac{P^{o} . X_{C H C l_{3}}}{P^{o} . X_{C H C l_{3}} + P^{o} . X_{C H_{2} C l_{2}}} = \frac{200 \times 0.5}{200 \times 0.5 + 415 \times 0.5} = 0.325$

Hence, the correct option is $A$

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Similar questions

Q. Vapour pressure of chloroform

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and dichloromethane

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(a) Vapour pressure of chloroform $(C H C l_{3})$ and dichloromethane $(C H_{2} C l_{2})$ at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of $C H_{2} C l_{2} a t 298 K$

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A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is:[Molar mass of C1=35.5 g mol−1]

A solution is prepared by mixing 8.5 g of $C H_{2} C l_{2}$ and 11.95 g of $C H C l_{3}$ . If vapour pressure of $C H_{2} C l_{2}$ and $C H C l_{3}$ at 298 K are 415 and 200 mmHg respectively, the mole fraction of $C H C l_{3}$ in vapour form is:

[Molar mass of C1=35.5 g $m o l^{- 1}$ ]