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Two liquids A and B on mixing form an ideal solution. Their vapour pressures in the pure state are 200 and 100 mm respectively. What will be mole fraction of B in the vapour phase in equilibrium with an equimolar solution of the two ?

Answer»

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Solution :In the liquid phase, `n_(B)=23.4//78=0.3, n_(T)=64.4//92=0.7." HENCE, "x_(B)=0.3, x_(T)=0.7`
`p_(B)=0.3xx75=22.5 mm, p_(T)=0.7xx22=15.4 mm , "Tota vapour PRESSURE "=22.5+15.4=37.9 mm"`
MOLE fraction of benzene in the vapour phase `=22.5//37.9=0.59.`


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