At `88^(o)C` benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction fo benzene in the mixture with toluene that will boil at `88^(o)C` at 1 atm pressure, benzene-toluene from an ideal solution ? (P of mixture = 760 torr)

Options

A. 0.416
B. 0.588
C. 0.68
D. 0.74

Answer

D

Explanation

`P_("mix")=760` torr because solution boils at `88^(@)C` Now
`P_(m)=P_("Benzene")^(@)X_("Benzene")^(@)+P_("toluene")^(@) X_("toluene")^(@)` `(X_("toluene")^(@)=1-X_("Benzene")^(@))`
`760 = 900 x m.f."of" C_(6)H_(6)+360 xx (1-m.f. "of" C_(6) H_(6))`
`"a is mol fraction of" C_(6)H_(6) "then" `
`:. 760 = 900a + 360 -360a,`
`:. a = 0.74`