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Standard XII

Chemistry

Elevation in Boiling Point

The solution ...

Question

The solution is prepared by mixing $8.5 g$ of $C H_{2} C l_{2}$ and $11.95 g$ of $C H C l_{3}$ at $298 K$ are $415$ and $200 m m H g$ respectively, the mole fraction of $C H C l_{3}$ in vapour form is: (Molar mass of $C l = 35.5 g m o l^{- 1}$ )
(JEE Main - 2017)

0.325

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0.486

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0.162

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0.675

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Solution

The correct option is A $0.325$
$Molar mass of C H C l_{3} = 119.5 g / m o l e$

$Molar mass of C H_{2} C l_{2} = 85 g / m o l e$

$Moles of C H C l_{3} = \frac{11.95}{119.5} = 0.1 m o l e$

$Moles of C H_{2} C l_{2} = \frac{8.5}{85} = 0.1 m o l e s$

$Mole fraction of CHCl3=0.10.2=0.5 mole$

$Mole fraction of CH2Cl2=0.10.2=0.5 mole$

Given:

$Vapour pressure of C H C l_{3} = 200 m m H g = 0.263 a t m$

$Vapour pressure of C H_{2} C l_{2} = 415 m m H g = 0.546 a t m$

$[∵ 1 a t m = 760 m m H g]$

$Pabove solution=Mole fraction of CHCl3×vapour pressure of CHCl3+Mole fraction of CH2Cl2×vapour pressure of CH2Cl2$

$= 0.5 \times 0.263 + 0.5 \times 0.546$

$= 0.4045$

$Mole fraction of CHCl3 in vapour form=0.13150.4045=0.325$

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Similar questions

Q. 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}

]

Q. Vapour pressure of chloroform

(C H C l_{3})

and dichloromethane

(C H_{2} C l_{2})

298 K are 200 m m H g and 415 m m H g

respectively. Calculate:

A

:Vapour pressure of the solution prepared by mixing

25.5 g o f C H C l_{3} and 40 g of C H_{2} C l_{2} at 298 K

B

: Mole fractions of each component in vapour phase.

Q. The vapour pressure of chloroform

({C H C l}_{3})

and dichloromethane

({C H}_{2} {C l}_{2})

at 298 K is 200 mm Hg and 415 mm Hg, respectively. If a solution is prepared by mixing 25.5 g of

{C H C l}_{3}

and 40 g of

{C H}_{2} {C l}_{2}

at 298 K, calculate the mole fractions of chloroform in vapour phase.

Q. Vapour pressure of chloroform

(C H C l_{3})

and dichloromethane

(C H_{2} C l_{2})

25^{o} C

are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of

C H C l_{3}

and 40 g of

C H_{2} C l_{2}

at the same temperature will be:

[Molecular mass of

C H C l_{3} = 119.5 g / m o l

and molecular mass of

C H_{2} C l_{2} = 85 g / m o l

]

Q. Vapour pressure of chloroform

(C H C l_{3})

and dichloromethane

(C H_{2} C l_{2})

25^{o} C

are

200 m m H g and 41.5 m m H g

respectively. Vapour pressure of the solution obtained by mixing

25.5 g

C H C l_{3}

and

40 g

C H_{2} C l_{2}

at the same temperature will be:
(Molecular mass of

C H C l_{3} = 119.5 u

and molecular mass of

C H_{2} C l_{2} = 85 u

)

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The solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1) (JEE Main - 2017)

The solution is prepared by mixing $8.5 g$ of $C H_{2} C l_{2}$ and $11.95 g$ of $C H C l_{3}$ at $298 K$ are $415$ and $200 m m H g$ respectively, the mole fraction of $C H C l_{3}$ in vapour form is: (Molar mass of $C l = 35.5 g m o l^{- 1}$ )
(JEE Main - 2017)