Home
Class 12
CHEMISTRY
1 mole of liquid A and 9 moles of liquid...

1 mole of liquid A and 9 moles of liquid B are mixed to form a solution. If `"P"_("B")^(@)=400"mm"` of Hg and `"P"_("A")^(@)=20 "mm"` of Hg at a temperature 'T' and normal boiling point of liquid B is 300K then answer the questions that follow.
Given data: `"K"_("b")=2.7"K kg mol"^(-1),` Molar mass of B=100
If 'A' is perfectly non volatile and it dimerises to and extent of 60% then what will be the vapour pressure of the solution.

A

`360mm Hg`

B

`3600/9.7" mm Hg"`

C

`4000/9.7"mm Hg"`

D

`36mm Hg`

Text Solution

Verified by Experts

The correct Answer is:
B

`{:(2A,rightarrow,"A"_(2),),(1,,0,),(1-beta,,beta//2,),(beta=0.6=0.4,,0.3,"Total=0.7):}`
`"P"_("S")="P"^(@)xx"solvent"`
`=400xx9/(9+0.7)`
`"P"_("S")=3600/9.7"mm of Hg"`
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

1 mole of liquid A and 9 moles of liquid B are mixed to form a solution. If "P"_("B")^(@)=400"mm" of Hg and "P"_("A")^(@)=20 "mm" of Hg at a temperature 'T' and normal boiling point of liquid B is 300K then answer the questions that follow. Given data: "K"_("b")=2.7"K kg mol"^(-1), Molar mass of B=100 If 'A' is assumed to be perfectly non volatile then what will be normal boiling point of the solution.

1 mole of liquid A and 9 moles of liquid B are mixed to form a solution. If "P"_("B")^(@)=400"mm" of Hg and "P"_("A")^(@)=20 "mm" of Hg at a temperature 'T' and normal boiling point of liquid B is 300K then answer the questions that follow. Given data: "K"_("b")=2.7"K kg mol"^(-1), Molar mass of B=100 It is observed that pressure of vapour above the solution at 'T' Kelvin is 350 mm Hg. The true statement is :

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.

1 mole heptane (V.P = 92 mm of Hg) is mixed with 4 mol. Octane (V.P = 31 mm of Hg) , form an ideal solution. Find out the vapour pressure of solution.

In binary ideal solution forms by liquid A and B, at constant temperature, mole -fraction of liquid A in vapour state is 0.4 and its partial vapour pressure is 400 mm,then what will be the partial vapour pressure of B ?

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

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

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

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-