A & B are two completely Miscible Liquids where A is solute & B is solvent.

A has 10% solubility in another solvent C. B & C r completely Immiscible.

So what you're saying is...that A will dissolve 1/10 as much in C as in B. (Assuming A is 100% soluble in B - because otherwise, it wouldn't be the solute any more, would it?)

Now if I start on 100 Kg solution basis of A & B with 20% concentration of A in B.

100 Kg of solution will have 20 Kg A & 80 Kg B.

Agreed.

Now I add 200 Kg of C in the Solution & mix it thoroughly and then allow to separate out 2 phases.

So now you have 20 kg of A dissolve among 80 kg of B and 200 kg of C.

Question is

What will be the amount of A in B & in C?

C dissolves A one tenth as much as B does. So you need 10 kg of C to dissolve as much A as 1 kg of B. In effect, the 200 kg of C is going to act just like 20 kg of B (20 is one tenth of 200).

Now it's like you have 20 kg of A dissolved among two systems: one with 80 kg of B, one with 20 kg of B. And the concentration of A is going to be the same in each of them...since they're both B (or at least, effectively). So, 80/(80+20) of the A will be dissolved in the REAL B, and 20/(80+20) will be dissolved in the pseudo-B (which is really C).

Hope that helps...maybe it was too confusing...I think it's alright.

I know it depends on Partition Coeff. of A in B & C but that is unknown at this stage.

I think I used partition coefficients without knowing it

And the whole second question can be solved similarly to the first one....and sometimes using a variable like x really really helps I find.

Man I hope I'm right on this...