Home
Class 12
CHEMISTRY
The vapour pressure of water at room tem...

The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to

A

`2.39 mm Hg`

B

`2.42 mm Hg`

C

`21.42 mm Hg`

D

`21.44 mm Hg`

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaP = p^(@) xx x_(B) = 23.8 xx 0.1 = 2.38`
`:.` Vapour pressure of sol `= 23.8 - 2.38`
`= 21.42 mm Hg`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    DINESH PUBLICATION|Exercise REVISION QUESTION|186 Videos

  • SOLUTIONS

    DINESH PUBLICATION|Exercise OBJECTIVE TYPE MCQs|47 Videos

  • SOLID STATE

    DINESH PUBLICATION|Exercise Brain storming|10 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos

Similar Questions

Explore conceptually related problems

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of water in a solution of non volatile solute is 0.9, the vapour pressure of the solution will be :-

The vapour pressure of water is 17.54mm Hg at 293 K. Calculate vapour pressure of 0.5 molal solution of a solute in it.

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be :

Benzene and toluene from an ideal solution. The vapour pressure of benzene at 55^(@) C is 400 mm Hg while the vapour pressure of toluene at 55^(@)C is 130 mm Hg. What is the vapour pressure of a solution consisiting of 0.5 mole fraction of benzene and 0.5 mole fraction of toluene at 55^(@)C ?

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 . Mole fraction of A in the solution , if it obeys Raoult's law is :