A solute has partition coefficient $K = C_{e t h er} / C_{w a t er} = 10$ . A $100 mL$ aqueous solution contains $5.0 g$ of the solute. It is extracted twice with $50 mL$ portions of ether. The mass of solute left in the aqueous layer is about

Answer: A

After one extraction the fraction left is

1/ (1 + K V_{e} / V_{w}) = 1/ (1 + 10 \times 50/100) = 1/6

. After two extractions the fraction left is

(1/6)^{2}

, so mass

= 5.0/36 = 0.139

g. A single

100

mL extraction would leave

0.455

g, so repeated small portions are better.