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Example:

The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure for solution and the mole fraction of methanol in the vapour.

Solution:

Mol. mass of ethyl alcohol = C2H2OH = 46

No. of moles of ethyl alcohol = 60/46 = 1.304

Mol. mass of methyl alcohol = CH3OH = 32

No. of moles of methyl alcohol = 40/32 = 1.25

‘XA’, mole fraction of ethyl alcohol = 1.304/1.304+1.25 = 0.5107

‘XB’, mole fraction of methyl alcohol = 1.25/1.304+1.25

Partial pressure of ethyl alcohol = XA. pA0 = 0.5107 × 44.5

= 22.73 mm Hg

Partial pressure of methyl alcohol = XB.pA0 =0.4893 × 88.7

= 43.73 m Hg

Total vapour pressure of solution = 22.73 + 43.40

= 66.13 mm Hg

Mole fraction of methyl alcohol in the vapour

= Partial pressure of CH3OH/Total vapour pressure = 43.40/66.13 = 0.6563

Example :

Two liquids A and B form ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.

Solution:

Let the vapour pressure of pure A be = pA0; and the vapour pressure of pure B be = pB0.

Total vapour pressure of solution (1 mole fraction of A and XB is mole fraction of B)

550 = 1/4 pA0 + 3/4 pB0

or 2200 = pA0 + 3pB0              …(i)

Total vapour pressure of solution (1 mole A + 3 moles B)

= 1/5 pA0 + 4/5 pB0

560 = 1/5 pA0 + 4/5 pB0

or 2800 = pA0 + 4pB0                                 …(ii)

Solving Eqns. (i) and (ii),

pB0 = 600 mm of Hg = vapour pressure of pure B

pA0 = 400 mm of Hg = vapour pressure of pure A

Example:

An aqueous solution containing 28% by mass of a liquid A (mo. Mass = 140) has a vapour pressure of 160 mm at 37oC. Find the vapour pressure of the pure liquid A. (The vapour pressure of water at 37oC is 150 mm).

Solution:

For total miscible liquids,

Ptotal = Mol. fraction of A × pA0 + Mol. fraction of B × pB0

No. of moles of A = 28/140

Liquid B is water. its mass is (100−28), i.e., 72.

No. of moles of B = 72/18

Total number of moles = 0.2 + 4.0 = 4.2

Given, Ptotal = 160 mm

pB0 = 150 mm

So 160 = 0.2/4.2 × pB0 + 4.0/4.2 × 150

pA0 = 17.15×4.2/0.2 = 360.15 mm

Example:

The vapour pressure of two pressure liquids A and B that forms an ideal solution at 300 and 800 torr respectively at temperature T. A mixture of the vapours of A and B for which the mole fraction A is 0.25 is slowly compressed at temperature T. Calculate:

(a) The composition of the first drop of condensation.

(b) The total pressure of this drop formed.

(c) The composition of the solution whose normal boiling point is T.

(d) The pressure when only the last bubble of vapours remains.

(e) Composition of the last bubble.

Solution:

(a) pB0 = 300 torr pB0 = 800 torr

yA = pA/P = pA0xA/P

where yA and yB are mole fraction of A and B in the vapour

yB = pB0xA/P

∴ yA = yB = pB0xA/pB0xB

0.025/0.75 = 300xA/800xB

i.e.,  xA/xB = 8/9

i.e.,   xA = 8/17 = 0.47
Composition of last drop
xB = 9/17 = 0.53

(B)    Pressure of last drop:

p = pA0xA + pB0xB

(c)    At boiling point: p = 760

760o = 300 xA + 800 (1−xA)

xA = 0.08,           xB = 0.92

(d)    For last drop:      xA = 0.25, xB = 0.75

P = 300 × 0.25 + 800 × 0.75 = 675 torr.

(e)    yA = pA/P

yA = 300×0.25/675 = 0.111

yB = 0.889

Here, yA and yB are composition of vapour of last drop.

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