Distribution Law

### Solution )

Suppose amount of acid extracted by ether  = x gm

Amount of acid left in water =( 5 – x )

Distribution coefficient , K = CW / CE

CW= [ (5 – x )  / 100]  gm. / 100ml

Given ,CE  = [ x  / 50]  gm /50ml

K = CW / CE

K = 5.5

5.5 = [(5 – x )  / 100] / [ x  / 50]

5.5 = 5 – x /2 x

x = 0.4167

x = 0.42 gm

### Solution )

Suppose , amount of acid in ether’ CE’ = x gm.

Amount of acid left in water ‘ CW’ = (7 – x ) gm.

K = CW / CE

K = 5.5

5.5 = (7 – x ) /  x

5.5 = 7 – x / x

5.5 x = 7 – x

6.5 x = 7

x = 1.0769 gm

### Solution )

K = 10

Suppose, amount of oxalic acid extracted by ether layer = [y / 100 ] gm / ml

Amount of oxalic aid left in water layer = ( 10 – y )/ 1000 gm / ml

K = CW / CE

K = 10.0

10  = [ (10 – y  )/ 1000] /  [y / 100]

10 =  10- y / 10 y

100 y  = 10 – y

101 y  = 10

y = 0.099 gm

### Solution )

Suppose, x and y are the volumes of  H2O and CCl4 respectively

Amount of iodine taken = 10 gm

Amount of iodine in water  layer = 5  gm

CWater = (5 / x ) gm / litre

Amount of iodine in left in ether  layer = (10 – 5 ) = 5  gm

CCCl4 = (5 / y ) gm / litre

According to Distribution Law,

K = CWater / CCCl4

K = 85

85 = [ (5/y )] /  [5 / x]

85 =  x / y

Ratio between volumes of  H2O and CCl4 = 85 : 1 Ans.