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Standard XII

Chemistry

Raoult's Law

The vapour pr...

Question

The vapour pressure of pure benzene and toluene at a particular temperature are $100 m m$ and $50 m m$ respectively. Then the mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is :

0.67

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0.75

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0.33

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0.50

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0.20

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Solution

The correct option is A $0.67$
Total vapour pressure, $p_{t o t a l} = p_{1} + p_{2}$
$= 100 + 50$
$= 150 m m$

$∴$ Mole fraction of benzene in solution,

$x = \frac{p_{1}}{p_{1} + p_{2}}$

$= \frac{100}{150} = \frac{2}{3} = 0.67$

Therefore, the correct option is A.

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The vapour pressure of pure benzene and toluene at a particular temperature are 100 mm and 50 mm respectively. Then the mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is :

The correct option is A 0.67Total vapour pressure, ptotal=p1+p2=100+50=150 mm∴ Mole fraction of benzene in solution,x=p1p1+p2=100150=23=0.67Therefore, the correct option is A.

The vapour pressure of pure benzene and toluene at a particular temperature are $100 m m$ and $50 m m$ respectively. Then the mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is :

The correct option is A $0.67$
Total vapour pressure, $p_{t o t a l} = p_{1} + p_{2}$
$= 100 + 50$
$= 150 m m$

$∴$ Mole fraction of benzene in solution,

$x = \frac{p_{1}}{p_{1} + p_{2}}$

$= \frac{100}{150} = \frac{2}{3} = 0.67$

Therefore, the correct option is A.