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Byju's Answer

Standard XII

Chemistry

Relative Lowering of Vapour Pressure

The vapour pr...

Question

The vapour pressure of chloroforms and de-chloroforms at 298 k are 200 mm Hg and 415 mm Hg? calculate the vapour pressure of solution by mixing 25.5 gram of $C H C l_{3}$ 40 gram of $C H_{2} C l_{2}$ . calculate the mole fraction of each component on vapour phase.

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Solution

$P_{A}^{0}$ (choloroforms) $= 200 m m H g$
$n_{A} = \frac{25.5}{(12 + 1 + 35.5 \times 3)} = 0.2134$
$P_{0}^{B} (C H_{2} C l_{2}) = 415 m m H g$
$n_{B} = \frac{40}{(12 + 2 + 31.5 \times 2)} = 0.4706$
$X_{A} = \frac{0.2134}{0.2134 + 0.4106 =} \frac{0.2134}{0.684}$
$= 0.312$
$X_{B} = 0.688$
$P_{A} = P_{A}^{o} X_{A} = 200 \times 0.312 = 62.4$
$P_{B} = P_{B}^{o} X_{B} = 0.685 \times 415 = 285.52$
$P = 62.4 + 285.52 = 347.92$
Mole fraction of $C H_{2} C l_{2}$ in vapour phase
$= \frac{285.52}{347.92} - 0.8206$
$≃ 0.82$
Mole fraction of $C H C l_{3}$ in vapour phase
$= \frac{62.4}{347.92} = 0.18$

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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})

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are

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and

415 m m H g

respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing

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and

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Q. Vapour pressure of chloroform

(C H C l_{3})

and dichloromethane

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

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and 40 g of

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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 $C H_{2} C l_{2} a t 298 K$