## Raoult’s Law

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### Solution )

po= 0.526 atm

Ps = 0.507 atm

po-ps =0.019

w = 0.869 gm

M of  ether (C4H10O)  = 12 x 4 + 1 x 10 + 1 x 16 = 74

m of cholesterol = ?

W= 4.4 gm.

(po-ps) /ps =wM/Wm

0.019 /0.507  =0.869 x 74 / 4.4 m

m = 0.869 x 74 x 0.507 /0.019 x 4.4

## m = 389.99  Ans.

### Solution )

T =373 K

Po of water = 1 atmosphere = 760 mm

Ps = 750 mm

M of water (H2O) = 1 x 2 + 1 x 16 = 18

According to Raoult’s Law

(po-ps) / po =wM / Wm

Molality = w x 1000/ m W

multiply by ‘M’ in numerator & denominator,

Molality = w x 1000 x M / m x W x M

because ,

(po-p) /po = wM/Wm

so,

Molality =  (po-p) x 1000 /po x M

= (760 – 750) x 1000 / 760 x 18

## Molality = 0.731 mole / Kg

mole fraction of solute ,

(n /n +N )= (po-p) /po

= (760 – 750)/760 = 10 / 760 =0.013

## mole fraction of solute = 0.013 Ans.

### Solution)

According to Raoult’s Law

Relative lowering in vapour pressure ,

po – ps / po = (n /n +N )

w =10 gm., m = 100, W = 180 gm. , M of water = 18

n = w/m = 10 / 100 =0.1

N = W/M= 180 /18 = 10

po – ps / po = 0.1 / (0.1 + 10)

= 0.1 /10.1

[ po – ps / p] = 0.0099

## Relative lowering in vapour pressure = 0.0099   Ans.

### Solution)

Relative lowering in vapour pressure,

po – ps / po = (n /n +N )

### For water,

w =6.0 gm., m = 120, W = 150 gm. , M of water = 18

n = w/m = 6 / 120 =0.0.05

N = W/M= 150 /18 = 8.33

po – ps / po = 0.05 / (0.05+ 8.33 )

### For ethyl alcohol ,

w =6.0 gm., m = 120, W = 150 gm. , M of  ethyl alcohol ( C2H5OH)  = 2 x 12 + 5 x 1 + 16 + 1 = 46

n = w/m = 6 / 120 =0.0.05

N = W/M= 150 /46 = 3.26

po – ps / po = 0.05 / (0.05+ 3.26 )

po – ps / po = 0.015