Question

You perform a single extraction with 100 mL of chloroform to remove compound A from 100 mL of an aqueous solution. The partition coefficient for compound A is 2.5. How much material remains in the aqueous solution after the extraction if you started with 5 g of compound A? Give your answer as a percent.

Answer 1

Answer:

Percentage of mass of solute in aqueous phase is 28.57%.

Explanation:

$(q)_{aq}$- Fraction of solute in aqueous phase.

$(q)_{aq}=\frac{V_{aqueous}}{DV_{organic}+V_{aqueous})}$

$V_{aqueous}$ = Used volume of aqueous solution

$V_{organic}$ = Used volume of organic solution

D = partition coefficient.

$(q)_{aq}=\frac{100ml}{(2.5\times100)+100\,ml}=0.2857$

Mass of solute in aqueous phase = $0.2857\times 5g=1.428g$

$Mass\,percentage = \frac{1.428}{5.0}\times 100=28.57$

Therefore, Percentage of mass of solute in aqueous phase is 28.57%.