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Given: mass of C_{6}H_{6} _{ }(w_{b})= 936 g

mass of C_{7}H_{8} (w_{a}) = 736 g

pressure of mixture, p = 180 X_{b }+ 90 mmHg

To calculate: mole fraction of benzene in vapour state

Solve: Mole fraction of benzene in liquid state (Xb) is given by

where n_{a }and n_{b} are number of moles of toluene and benzene respectively.

Number of moles is given by

where w is the given mass and M is molar mass of the substance.

Molar mass of Benzene (M_{b}) = 78 g mol^{-1}

Molar mass of Toluene (M_{a}) = 92 g mol^{-1}

nb = wb/Mb

n_{b} = 12 mol

na = wa/Ma

na = 8 mol

Xb = 0.6

and X_{a} = 1 - X_{b}

_{ } = 1 - 0.6

=0.4

Thus substituting the value of X_{b} in the given equation of pressure

p = 180 X_{b} _{ }+ 90 mmHg

p = 180 × 0.6 +90 mmHg

p = 198 mmHg

Comparing the given equation of pressure

p = (180 X_{b} _{ }+ 90 ) mmHg

with the standard equation of

thus,

mole fraction of benzene in the vapour state is given by

and

where is the partial pressure of pure benzene at 50 deg. C.

p_{b}= 90 ×0.6

= 54 mmHg

Thus

*y**b*= 54/198 = **0****.****27**

Therefore, the mole fraction of benzene in vapour state is 0.27.

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( PS I had edited your question because the image you had added was too large. Please type the questions in such cases)