Benzene (l) $⇌$ Benzene (g)

$K_{P} = 60 m m H g$ at $300 K$

Toulene (l) $⇌$ Toulene (g):

$K_{P} = 40 m m H g$ at $300 K$

The mole fraction of benzene in vapours of benzene-toulene mixture is $0.6$ at $300 K$ temperature. The mole fraction of Toulene in liquid phase of solution at $300 K$ is____________.

0.5

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0.55

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0.60

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None of these

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Solution

The correct option is A $0.5$
The number of moles is the ratio of mass to molar mass.

The molar masses of benzene and toluene are 7878 g/mol and 9292 g/mol respectively.

Number of moles of benzene $= \frac{80}{78} = 1.026$

Number of moles of toluene $= \frac{100}{92} = 1.087$

Mole fraction of benzene, $X_{B} = \frac{1.026}{1.026 + 1.087} = 0.486 X$

Mole fraction of toluene, $X_{T} = 1 - 0.486 = 0.514 \approx 0.5$

Hence, the correct option is $A$

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Benzene (l)⇌ Benzene (g)KP=60mmHg at 300KToulene (l) ⇌ Toulene (g):KP=40mmHg at 300KThe mole fraction of benzene in vapours of benzene-toulene mixture is 0.6 at 300K temperature. The mole fraction of Toulene in liquid phase of solution at 300K is____________.