Home
Class 12
CHEMISTRY
Vapour pressure of mixture of liquid A a...

Vapour pressure of mixture of liquid A and liquid B at `70^(@)C` is given by `P_(T)=180X_(B)+90 (in mm)`. Where `X_(B)` is the mole fraction of B, in the liquid mixture. Calculate
(a) Vapour pressure of pure A and pure B
(b) Vapour pressure of mixture of A and B by mixing 4 g and 12 g B. (If molar mass of A and B are 2 g and 3 g respectively)
(c ) From (b) ratio of moles of A and B in vapour at `70^(@)C`

Text Solution

Verified by Experts

(a) 90 mm Hg 270 mm Hg
(b) 210 mm Hg
(c ) `1:6`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGMENT (SECTION-J) AAKASH CHALLENGERS QUESTIONS|10 Videos

  • SOLUTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGMENT (SECTION-G) TRUE-FALSE TYPE QUESTIONS|5 Videos

  • REDOX REACTIONS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - D|10 Videos

  • SOME BASIC CONCEPT OF CHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT( SECTION - D) Assertion-Reason Type Questions|15 Videos

Similar Questions

Explore conceptually related problems

The vapour pressures of pure liquid A and pure liquid B at 20^(@)C are 22 and 75 mm of Hg respectively. A solution is prepared by mixing equal moles of A and B. Assuming the solution to be ideal, calculate the vapour pressure of the solution.

In a mixture of A and B having vapour pressure of pure A and B are 400 m Hg and 600 mm Hg respectively. Mole fraction of B in liquid phase is 0.5. Calculate total vapour pressure and mole fraction of A and B in vapour phase .

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:

If P_(A) is the vapour pressure of a pure liquid A and the mole fraction of A in the mixture of two liquids A and B is x, the parial vapour pressure of A is:

The vapour pressure of pure liquid A and liquid B at 350 K are 440 mm and 720 mm of Hg. If total vapour pressure of solution is 580 mm of Hg then the mole fraction of liquid A in vapour phase will be :-

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 liquids A and B is 450 and 700mm Hg , respectively, at 350K. Find out the composition of the liquid mixture if the total vapour pressure is 600mm Hg . Also find the composition of the vapour phase.

At a given temperature total vapour pressure (in Torr) of a mixture of volatile components A and B is given by P = 120 - 75 X_(B) hence vapour pressure of pure A and B respectively (in Torr) are

Vapour pressure of pure liquids P and Q are 700 and 450 mm Hg respectively at 330K. What is the composition of the liquid mixture at 330 K, if the total vapour pressure is 600 mm Hg?

The total vapour pressure of a mixture of 1 mole of A ( P_A = 200 mm Hg ) and 3 mole of B ( P_B= 360 mm Hg) is 350 mm Hg. Then