Distribution Law

source :Chromatography-online.org

Numericals of Distribution Law

### Ans.

k1 =Cwater / Cether

=   0.0244/.0046

k=5.30

k2 = Cwater /  Cether

=0.071 /0.013

k2= 5,46

k3 = Cwater /  Cether

=0.121 /0.022

k= 5.5

k =( k1+k2+k3 ) /3

= (5.30 +5.46 +5.5 )/3

=16.26/3

## K=5.42

This data follows the distribution law because value of K, K2 & K3 are nearly equal

### K1 = Cwater /  Cether

=25.4 / 4.2 =6.05

K2 = 33,2/5.5 =6.04

K3 = 42.6 / 7.1 =6.00

## The above results are in favour of distribution law because values of K1 , K2 & K3 are nearly equal.

### Q.3      Iodine (5.0 gm) was dissolved in a mixture containing equal volumes of water & CCl4. If K of iodine between CCl4 & H2O is 82, find the amount of I2 in aqueous layer?

Ans.

Suppose,

Amount of I2 in aqueous layer = x

Amount of I2 in aqueous CCl4 layer =5- x

K =C CCl4 / C H2O

82  = ( 5-x) /x

82 x =5-x

83 x =5

x=5/83 =0.0602  Ans.

### Ans.

SWater               = 0.8 gm/l

SCCl4     =          ? gm/l

K=  SCCI4  / Swater

82 = SCCI4 /0.8

SCCI4  =82 X0.8 =65.6 gm. / l   Ans.

### Solution )

Distribution coefficient , K = CW / CE

K = 5.2

CW = x  gm / 25 ml

Given ,CE  = 0.046 gm / 10 ml

CE  = 0.046 x 25 / 10 = 0.115 gm / 25 ml

K = CW / CE

5.2 = x / 0.115

x = 5.2 x 0.115= 0.598