2 liquids A and B form ideal solution . one mole of A and B both are mixed to form a solution . now 1 mole of the mixture has been vapourise. calculate vapour pressure of the solution at this point. P0A- 900 torr P0B=100 Torr

AS PER QUESTION, 1 MOLE OF THE MIXTURE HAS BEEN VAPOURISE . WE HAVE TO FIND THE COMPOSITION OF BOTH LIQUID IN VAPOUR MIXTURE. P=PRESSURE OF MIXTURE

NOW USING , PV=nRT

(POA.XA+POB.XB).V=nRT

(900*.5+100*.5).V=2RT = V/RT =2/5000=4*10^-4.

P=REQUIRED PRESSURE , N=1 therefore , p=NRT/V=N/(V/RT)=1/(4*10^-4)=2500.

Previous answer is wrong the correct one is this..As let Yb be mole fraction of B in gasAnd Xb in liq form Yb=nb/1Xb=(1-nb)/1PTotal=P°aXa+P°bXbAlso P°bXb=YbPTotalNow substitute the valuesYou will get Ptotal=√(P°a*P°b)Ptotal=300 torr