Home
Class 12
CHEMISTRY
A certain ideal solution of two liquids ...

A certain ideal solution of two liquids `A` and `B` has mole fraction of `0.3` and `0.5` for the vapour phase and liquid phase, respectively. What would be the mole fraction of `B` in the vapour phase, when the mole fraction of `A` in the liquid is `0.25`?

Text Solution

AI Generated Solution

To solve the problem, we need to determine the mole fraction of component B in the vapor phase when the mole fraction of component A in the liquid phase is given as 0.25. ### Step-by-Step Solution: 1. **Identify Given Values**: - Mole fraction of A in the vapor phase, \( y_A = 0.3 \) - Mole fraction of B in the liquid phase, \( x_B = 0.5 \) - Mole fraction of A in the liquid phase, \( x_A = 0.25 \) ...
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Examples|40 Videos
  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Linked Comprehension)|58 Videos
  • SOLID STATE

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.2 (Objective)|9 Videos
  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|2 Videos

Similar Questions

Explore conceptually related problems

If Pa and Pb are 108 and 36 torr respectively. What will be the mole fraction of A is vapour phase if B has mole fraction in solution 0.5:

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 phese are 1 : 3 then mole fraction of 'A' in vapour phase in equilibrium with the solution is equal to :

At 300 K two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In a equimolar liquid mixture of A and B, the mole fraction of B in the vapour phase above the solution at this temperature is:

The vapour pressure of a pure liquid at 25^(@)C is 100 mm Hg. Calculate the relative lowering of vapour pressure if the mole fraction of solvent in solution is 0.8.

The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is:

At a certain temperature, the vapour pressure of pure ether is 640 mm and that of pure acetone is 280 mm . Calculate the mole fraction of each component in the vapour state if the mole fraction of ether in the solution is 0.50.

Two liquids A and B have vapour pressure of 0.600 bar and 0.2 bar, respectively. In an ideal solution of the two, calculate the mole fraction of A at which the two liquids have equal partial pressures.

Two liquid A and B have vapour pressure in the ratio P_A^(@) : P_B^(@) =1.3 at a certain temperature.Assume A and B from an ideal solution and the ratio of mole fractions of A to B in the vapour phase is 4 : 3, then the mole fraction of B in the solution at the same tempreature is : (a) 1/5 (b) 2/3 (c) 4/5 (d) 1/4

If the mole fraction of Iodine in benzene is 0.25, then what will be the molality of the solution ?

If the vapour pressure of pure A and pure B at 298 K are 60 and 15 torr respectively, what would be the mole fraction of A in vapour phase (at this temperature) in a solution that contains 20 mole per cent of A in the (A + B) binary mixture in liquid phase ?