Answer:
0.00370 g
Explanation:
From the given information:
To determine the amount of acid remaining using the formula:[tex]\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n[/tex]
where;
v_1 = volume of organic solvent = 20-mL
n = numbers of extractions = 4
v_2 = actual volume of water = 100-mL
k_d = distribution coefficient = 10
∴
[tex]\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4[/tex]
[tex]\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4[/tex]
[tex]\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4[/tex]
[tex]\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345[/tex]
Thus, the final amount of acid left in the water = 0.012345 * 0.30
= 0.00370 g
