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Vapour pressu...

Question

Vapour pressure of chloroform $(C H C l_{3})$ and dichloromethane $(C H_{2} C l_{2})$ at $298 K are 200 m m H g and 415 m m H g$ 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.

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Solution

$A$ :

Given:

Vapour pressure of pure $C H C l_{3} (P_{C H C l_{3}}^{0}) = 200 m m H g$ ,

Vapour presssure of pure $C H_{2} C l_{2} (P_{C H_{2} C l_{2}}^{0}) = 415 m m H g$ ,

Mass of $C H C l_{3} = 25.5 g and mass of C H_{2} C l_{2} = 40 g$

Moles of compounds

Molar mass of $C H_{2} C l_{2} = 12 \times 1 + 1 \times 2 + 35.5 \times 2 = 85 g m o l^{- 1}$

Molar mass of $C H C l_{3} = 12 \times 1 + 1 \times 35.5 \times 3 = 119.5 g m o l^{- 1}$

Moles of $C H_{2} C l_{2} = \frac{mass}{molar mass}$
$= \frac{40 g}{85 g m o l^{- 1}}$
$= 0.47 m o l$
Similarly,

Moles of $C H C l_{3} = \frac{25.5 g}{119.5 g m o l^{- 1}}$
$= 0.213 m o l$

Mole fraction of compounds

We know,

Mole fraction $(x) = \frac{moles of one component}{total moles of all the component}$
Total number of moles $= 0.47 + 0.213 = 0.683 m o l$

$x_{C H_{2} C l_{2}} = \frac{0.47 m o l}{0.683 m o l} = 0.688$

$x_{C H_{2} C l_{2}} + x_{C H C l_{3}} = 1$ ... $(i)$

$x_{C H C l_{3}} = 1.00 - 0.688 = 0.312$

Total pressure of mixture

We know,
$P_{total} = P_{C H C l_{3}} + P_{C H_{2} C l_{2}}$

And $P_{gas} = P_{gas}^{0} \times x_{gas}$

So, $P_{total} = P_{C H C l_{3}}^{0} \times x_{C H C l_{3}} + P_{C H_{2} C l_{2}}^{0} \times x_{C H_{2} C l_{2}}$

$P_{total} = P_{C H C l_{3}}^{0} \times (1 - x_{C H_{2} C l_{2}}) + P_{C H_{2} C l_{2}}^{0} \times x_{C H_{2} C l_{2}}$

$from equation (i)$

$p_{total} = p_{C H C l_{3}}^{0} + (p_{C H_{2} C l_{2}}^{0} - p_{C H C l_{3}}^{0}) x_{C H_{2} C l_{2}}$

By putting the values, we get,
$= 200 + (415 - 200) \times 0.688$

$p_{total} = 200 + 147.92 = 347.92 m m H g$

Given:

From the first part,

$P_{total} = 347.92 m m H g$ ,

Mole fraction of $C H_{2} C l_{2} = 0.688$ ,

Mole fraction of $C H C l_{3} = 0.312$ ,

Vapour pressure of pure $C H C l_{3} (P_{C H C l_{3}}^{0}) = 200 m m H g$ ,

And vapour pressure of pure

$C H_{2} C l_{2} (P_{C H_{2} C l_{2}}^{0}) = 415 M M H g$ ,

Mole fraction in gas phase

We know,

Mole fraction in gas phase $y_{gas} = P_{gas} / P_{total}$

From Raoult’s law, $P_{C H_{2} C l_{2}} = P_{C H_{2} C l_{2}}^{0} \times x_{C H_{2} C l_{2}}$

$P_{C H_{2} C l_{2}} = 415 M M H g \times 0.688$
$= 285.52 m m H g$

Similarly,

$P_{C H C l_{3}} = 0.312 \times 200 m m H g = 62.4 M M H g$

So, $Y_{C H_{2} C l_{2}} = \frac{P_{C H_{2} C l_{2}}}{P_{t o t a l}} = \frac{285.52 M M H g}{347.92 m m H g}$
$= 0.82$

Similarly,

$Y_{C H C l_{3}} = \frac{62.4 m m H g}{347.92 m m H g} = 0.18$

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Similar questions

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}

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

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

Q. 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}

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. calculate the mole fraction of each component on vapour phase.

Q. Vapour pressure of chloroform

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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

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C H_{2} C l_{2}

at the same temperature will be:

[Molecular mass of

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and molecular mass of

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]

Q. Vapour pressure of chloroform

(C H C l_{3})

and dichloromethane

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298 K

200 m m H g

415 m m H g

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

25.5 g

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40 g

C H_{2} C l_{2}

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and dichloromethane

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

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200 m m H g and 41.5 m m H g

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