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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.

Respuesta :

Answer:

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

Explanation:

[tex](q)_{aq}[/tex]- Fraction of solute in aqueous phase.

[tex](q)_{aq}=\frac{V_{aqueous}}{DV_{organic}+V_{aqueous})}[/tex]

[tex]V_{aqueous}[/tex] = Used volume of aqueous solution

[tex]V_{organic}[/tex] = Used volume of organic solution

D = partition coefficient.

[tex](q)_{aq}=\frac{100ml}{(2.5\times100)+100\,ml}=0.2857[/tex]

Mass of solute in aqueous phase = [tex]0.2857\times 5g=1.428g[/tex]

[tex]Mass\,percentage = \frac{1.428}{5.0}\times 100=28.57[/tex]

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

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