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:
where, pB=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.
pT=total pressure in the vapour phase = (100+200) mm Hg = 300 mm Hg
xB=mole fraction of B.
xB=pB/pT = 100 / 300 = 0.33