Solution :
The partition coefficient

= 9.0
A). 1 x 200 mL extraction
Let m be the mass of caffeine in water
Mass of caffeine in
= 100 - m
∴




= 10
Therefore, the mass remaining in the coffee is m = 10 mg
B). 2 x 100 mL extraction
First extraction :
Let
be the mass of the caffeine in water.
Mass of caffeine in
= 100 - m
∴




= 18.18
Mass remaining in the coffee after the 1st extraction
= 18.18 mg
Second extraction:
Let
be the mass of the caffeine in water.
Mass of caffeine in
= 18.18 -

∴




= 3.3
Mass remaining in the coffee after the 1st extraction
= 3.3 mg