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

*The answer is 0.33..!!! how to get it??*

We know that, mole fraction can be calculated using the following formula:

p_{B}=p_{T}*x_{B}.

where, p_{B}=partial pressure of B = 100 mm Hg as in the vapour phase the partial pressure of B will be equal to its vapour pressure in poure state.

p_{T}=total pressure in the vapour phase = (100+200) mm Hg = 300 mm Hg

x_{B}=mole fraction of B.

x_{B}=p_{B}/p_{T} = 100 / 300 = 0.33

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