Two liquids A and B on mixing form ideal solution their vapour pressures in pure state are 200 and - Chemistry - Solutions

Question

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??

Kanika Dalal · Accepted Answer

We know that, mole fraction can be calculated using the
following formula:
pB=pT*xB
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

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??

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??