Question

What weight of “A” can be removed from a solution that contains 6.00 grams of “A” in 50.0 mLs of water by a single extraction using 50.0 mLs of ligroin?

Answer 1

Hey there!:

Kd for "A" = C (ligroin) / C (water) = 8.0

6 gm A in 50 mL water using 50 mL ligorin in single extraction.

Let x g of A is extracted.

Kd = (x / 50 mL ligorin) / ((6 - x) / 50 mL water)

8.0 = x / (6 - x)

48 - 8*x = x

9*x = 48

x = 48 / 9

x = 5.33 g

Hope that helps!