Home
Class 12
CHEMISTRY
The vapour pressure of a pure liquid A i...

The vapour pressure of a pure liquid `A` is `40 mm Hg` at `310 K`. The vapour pressure of this liquid in a solution with liquid `B` is `32 mm Hg`. The mole fraction of `A` in the solution, if it obeys Raoult's law, is:

A

0.8

B

0.5

C

0.2

D

0.4

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(A)=40 mm Hg`,`P_(A)=P_(A)^(@)chi_(A)=32 rArr chi_(A) = 0.8`,`chi_(B)=0.2`
Promotional Banner

Similar Questions

Explore conceptually related problems

The vapour pressure of methyl alcohol at 298 K is 0.158 bar. The vapour pressure of this liquid in solution with liquid B is 0.095 bar. Calculate the mole fraction of methyl alcohol in the solution if the mixture obeys Raoult's law.

The vapour pressure of pure benzene is 639.7 mm Hg and the vapour pressure of solution of a solute in benzene at the temperature is 631.9 mm Hg . Calculate the molality of the solution.

The vapour pressure of a certain pure liquid A at 298 K is 40 mbar. When a solution of B is prepared in A at the same temperature, the vapour pressure is found to be 32 mbar. The mole fraction of A in the solution is

The vapour pressure of pure liquid solvent A is 0.80 atm . When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm , the mole fraction of component B in the solution is

Why is the vapour pressure of liquid constant at a constant temperature ?

Calculate the vapour pressure of an aqueous solution which contains 5 mass per per cent of urea . The vapour pressure of pure water is 23.5 mm Hg . The molar mass of urea is 60 .

At 20^@C , the vapour pressure of pure liquid A is 22 mm Hg and that of pure liquid B is 75 mm Hg . What is the composition of the solution of these two components that has vapour pressure of 48.5 mm Hg at this temperature?