Home
Class 12
CHEMISTRY
What will be the relative lowering in va...

What will be the relative lowering in vapour pressure when 1 mole of A(l) is dissolved in 9 moles of B(l). Given both are volatile with vapour pressures as 100 mm of Hg and 300 mm of Hg respectively?

A

`0.1`

B

`0.067`

C

`20`

D

`30`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 10|12 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 11|12 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 1|23 Videos

  • IONIC EQUILIBRIUM

    GRB PUBLICATION|Exercise All Questions|526 Videos

  • METALLUGY

    GRB PUBLICATION|Exercise Subjective type|1 Videos

Similar Questions

Explore conceptually related problems

Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

The vapour pressure of water at 23^(@)C is 19.8 mm of Hg 0.1 mol of glucose is dissolved in 178.2g of water. What is the vapour pressure (in mm Hg) of the resultant solution?

When one mole of non-volatile solute is dissolved in three moles of solvent, the vapour pressure of the solution relative to the vapour pressure of the pure solvent is -

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 :

At 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg, if one more mole of B is added to the above ideal solution at 300K. Then the vapour pressure of A in its pure state is

Moles of Na_(2)SO_(4) to be dissoved in 12 mole water to lower its vapour pressure by 10 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is: