Home
Class 12
CHEMISTRY
A mixture contains 1 mole of volatile li...

A mixture contains 1 mole of volatile liquid A `("P"_("A")^(0)=80"mm Hg").` and 3 moles of volatile liquid B `("P"_("A")^(0)=80"mm Hg").` If solution behaves ideally, the total vapour pressure of the distillate is

A

25 mm Hg

B

85 mm Hg

C

90 mm Hg

D

92 mm Hg

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 38|4 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 40|4 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 25|8 Videos

  • IONIC EQUILIBRIUM

    GRB PUBLICATION|Exercise All Questions|526 Videos

  • METALLUGY

    GRB PUBLICATION|Exercise Subjective type|1 Videos

Similar Questions

Explore conceptually related problems

Total vapour pressure of mixture of 1 mole of volatile component A (P_(A)^(@)=100 mm Hg) and 3 mole of volatile component B (P_(B)^(@)=80 mm Hg) is 90 mm Hg . For such case :

Total vapour pressure of mixture of 1 mol of volatile component A( p_(A)^(@)=100 mm Hg) and 3 mol of volatile component B( p_(B)^(@) =60 mm Hg) is 75 mm. For such case:

Total vapour pressure of mixture of 1 mole of volaile components A ( P_(a^(%) )=100 mm Hg) and 3 mole of volatile component B( P_B^(@) =80 mm Hg ) is 90 mm Hg. For such case:

An ideal solution contains two volatile liquids A(P^(@)=100 torr) and B(P^(@)=200 torr). If mixture contain 1 mole of A and 4 mole of B then total vapour pressure of the distillate is :

Total vapour pressure of mixture of 1 mol volatile component A(P^(@)A="100 mm Hg") and 3 mol of volatile component B(P^(@)B="60 mm Hg") is 75 mm. For such case -

Liquids A (p_(A)^(@)=360" mm Hg") and B(p_(B)^(@)=320" mm Hg") are mixed. If solution has vapour has vapour pressure 340 mm Hg, then number of mole fraction of B/mole solution will be

Pressure over ideal binary liquid mixture containing 10 moles each of liquid A and B is gradually decreased isothermally. If "P"_("A")^(@)=200"mm Hg " "and"" P"_("B")^(@)=100"mm Hg, " find the pressure at which half of the liquid is converted into vapour.

If an ideal solution is made by mixing 2 moles of benzene (p^(@)=266mm) and 3 moles of another liquid (p^(@)=236 mm) . The total vapour pressure of the solution at the same temperature would be

Two liquids A and B are mixed to form an ideal solution. The totol vapour pressure of the solution is 800 mm Hg. Now the mole fractions of liquid A and B are interchanged and the total vapour pressure becomes 600 mm of Hg. Calculate the vapour pressure of A and B in pure form. (Given : p_(A)^(@)-p_(B)^(@)=100 )