Solve this:
Q. Two liquids A and B form ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increase by 10 mm Hg. The vapour pressure of A and B in their pure state (in mm Hg) respectively are
(1) 400, 600 (2) 500, 500
(3) 200, 400 (4) 300, 600
Dear student,
acc. to Raoult's law Pmix = XA. PA0 + XB . PB0
now XA = 1 /4 and XB = 3 /4 ; then in first condi. 550 = 1 /4 PA0 + 3 /4 PB0 -----(i)
again on adding 1 mole of B in the mix. XA = 1 /5 and XB = 4 /5 then (550 +10) = 1 /5 PA0 + 4 /5 PB0 ------(ii)
on solving the above eq. as follows ; 550 4 = PA0 + 3 PB0 ------(i)
560 5 = PA0 + 4 PB0 -------(ii)
then; PA0 = 400 mm Hg and PB0 = 600 mm Hg (option 1)
Regards
acc. to Raoult's law Pmix = XA. PA0 + XB . PB0
now XA = 1 /4 and XB = 3 /4 ; then in first condi. 550 = 1 /4 PA0 + 3 /4 PB0 -----(i)
again on adding 1 mole of B in the mix. XA = 1 /5 and XB = 4 /5 then (550 +10) = 1 /5 PA0 + 4 /5 PB0 ------(ii)
on solving the above eq. as follows ; 550 4 = PA0 + 3 PB0 ------(i)
560 5 = PA0 + 4 PB0 -------(ii)
then; PA0 = 400 mm Hg and PB0 = 600 mm Hg (option 1)
Regards