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

The vapour pressure of water at room temperature is lowered by `5%` by dissolving a solute in it , then the approximately molality of solution is :

A

2

B

1

C

4

D

3

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 33|7 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 36|6 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 30|7 Videos

  • IONIC EQUILIBRIUM

    GRB PUBLICATION|Exercise All Questions|526 Videos

  • METALLUGY

    GRB PUBLICATION|Exercise Subjective type|1 Videos

Similar Questions

Explore conceptually related problems

The vapour pressure of water at room temperature is lowered by 5% by dissolving a solute in it. What is approximate molality of solution:

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 :

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 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 :

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

How mich urea ("molar mass"=60 g mol^(-1)) must be dissolved in 50 g of water so that the vapour pressure at room temperature is reduced by 25% ? Also calculate the molality of the solution obtained.

If at certain temperature, the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous urea solution is 24.5 mm Hg, the molality of the solution is

Give two points to explain why vapour pressure of solvent is lowered by dissolving nonvolatile solute into it.