Two liquids A and B have `P_A^(@)" and P_B^(@)` in the ratio of 1 : 3 and the ratio of number of moles of A and B in liquid phase are 1 : 3 then mole fraction of 'A' in vapour phase in equilibrium with the solution is equal to :
A
0.1
B
0.2
C
0.5
D
1
Text Solution
Verified by Experts
The correct Answer is:
a
Topper's Solved these Questions
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 2|1 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
If two substance A and B have P_A: P_B=1 : 2 and also X_A : X_B in solutions as 1 : 2, then mole fraction of A in vapours is
If two substances A and B have p_A^0 : p_B^0 = 1: 2 and have mole fraction in solution 1: 2, then mole fraction of A in vapours is
Two liquids A and B have vapour pressure in the ratio P_(A)^(0):P_(B)^(0)=1:3 at a certain temperature . Assume A and B from an ideal solution and the ratio of mole fraction of A to B in the vapour phase is 4:3 . Then the mole fraction of B in the solution at the same temperature is :
Two liquids A and B are at 25^(@)C and 15^(@) C. Their masses are in 2 : 3 and specific heats in ratio 3 : 4 then resultant temperature when they are mixed is
The vapour presssures of pure benzene and toluene are 160 and 60 mm Hg respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is
Two liquids A and B form an Ideal solution. At 300K, the V.P of solution containing one mole of 'A' and 4 mole 'B' is 560mm Hg. At the same temp. If one mole of 'B' is taken out from the solution the V.P of the solution has decreased by 10mm Hg, the V.P, of pure A & B are (in min)
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column