Raoult’s law-

## Raoult’s law-

### Solution)

Initially,

PM =550 mm, PA0 = ? , PB0 =?, nA = 1 , nB = 3

PM = PA0.XA + PB0.XB

= PA0 [nA/(nA + nB)] +  PB0 [nB/(nA + nB)]

550 = PA0 [1/(1 + 3)] +  PB0 [3 /(1 + 3)]

550 = ( PA0/4 ) + ( 3PB0/ 4)

550 x 4 = PA0 + 3PB0

PA0 + 3PB0 =2200        ——– eq.1

On addition of one mole of B,

nA = 1 , nB = 3+1 = 4

The vapour pressure of solution increases by 10 mm Hg . Thus ,

PM =550 + 10 = 560 mm

PM = PA0.XA + PB0.XB

= PA0 [nA/(nA + nB)] +  PB0 [nB/(nA + nB)]

560 = PA0 [1/(1 + 4)] +  PB0 [4 /(1 + 4)]

560 = ( PA0/5 ) + ( 4 PB0/ 5)

560 x 5 = PA0 + 4PB0

PA0 + 4 PB0 =2800        ——– eq. 2

PA0 + 3 PB0 =2200

PA0 + 4 PB0 =2800

By Subtracting ,

– PB0 =-600

## PB0 = 600 mm Ans.

putting the value of PB0 in eq. 1

PA0 + 3 x 600 = 2200

PA0 = 2200 – 1800 = 400 mm

## PA0 = 400 mm Ans.

### Solution )

0.1 M solution of glucose is isotonic with solution of X in 100 gm water.

Therefore , concentration of solution of X = 0.1 M

It means 0.1 mole of X in 1000 ml or one litre of solution.

#### For dilute solution ,

volume of solution = volume of solvent (water) = weight of water( = v x d = 1000 x 1= 1000 gm)

W = 1000 gm , M of water = 18

mole of X  ‘n’= 0.1, mole of water  ‘N’ =W/M = 1000/18

According to Raoult’s law

po – ps / po = (n /n +N )

= 0.1 / [ 0.1 + (1000/18)]

## Relative lowering of vapour pressure = po – ps / po = 0.0018 Ans.

### Solution )

ps = 4/5 po , w = ? , m = 60 , W = 180 gm , M of water = 18

According to Raoult’s law

(po-ps) /po =wM/Wm

[po– (4p/5) /po] = w x 18 /180 x 60

(5 po – 4 po) /5 po = w /10 x 60

1 /5 = w / 600

w = 600 /5 = 120

## w = 120 gm.

### Solution )

PM = 290 mm , PoC2H5OH= ? ,

PM = PoC2H5OH.XoC2H5OH+ PoC3H7OH.XoC3H7OH

X C2H5OH = 0.65 , XC3H7OH = 1 –  0.65 = 0.35 , PoC3H7OH = 210 mm

290  = PoC2H5OH x 0.65 + 0.35 x 210

290  = PoC2H5OH x 0.65 + 73.5

290 – 73.5 = PoC2H5OH x 0.65

PoC2H5OH  = 216.5 / 0.65 = 333.07