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Standard XII

Chemistry

Raoult's Law

Mole fraction...

Question

Mole fraction of benzene in the vapours over a solution of benzene and toluene is 0.4. If the original vapour pressure of benzene is $150 m m H g$ and toluene is $50 m m H g$ , the mole fraction of benzene in the solution is:

0.4

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0.75

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0.25

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0.18

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Solution

The correct option is D 0.18
$T o t a l V P = 150 x_{B} + 50 x_{T}$
$= 150 x_{B} + 50 (1 - x_{B})$
$= 50 + 100 x_{B}$
$150 x_{B} = (50 + 10 x_{B}) \times 0.4$
$150 x_{B} = 20 + 40 x_{B} \Rightarrow x_{B} = \frac{20}{110} = 0.18$

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Mole fraction of benzene in the vapours over a solution of benzene and toluene is 0.4. If the original vapour pressure of benzene is 150 mm Hg and toluene is 50 mm Hg, the mole fraction of benzene in the solution is:

The correct option is D 0.18Total VP=150 xB+50 xT =150 xB+50(1−xB) =50+100 xB 150 xB=(50+10 xB)×0.4 150 xB=20+40 xB⇒ xB=20110=0.18

Mole fraction of benzene in the vapours over a solution of benzene and toluene is 0.4. If the original vapour pressure of benzene is $150 m m H g$ and toluene is $50 m m H g$ , the mole fraction of benzene in the solution is:

The correct option is D 0.18
$T o t a l V P = 150 x_{B} + 50 x_{T}$
$= 150 x_{B} + 50 (1 - x_{B})$
$= 50 + 100 x_{B}$
$150 x_{B} = (50 + 10 x_{B}) \times 0.4$
$150 x_{B} = 20 + 40 x_{B} \Rightarrow x_{B} = \frac{20}{110} = 0.18$