# Question 26fdf

Apr 24, 2015

The answer is (4) $\text{124/125 g}$

Here's how the problem looks like

The key to this problem is the distribution coefficient, or ${K}_{D}$, because it will tell you the ratio that exists between the concentration of A in carbon tetrachloride, $C C {l}_{4}$, and in water.

Since you know that $\text{A}$ is more soluble in $C C {l}_{4}$ than it is in water, you can write

${K}_{D} = \frac{{\left[A\right]}_{C C {l}_{4}}}{{\left[A\right]}_{{H}_{2} O}} = 4$, or ${K}_{D}^{- 1} = \frac{{\left[A\right]}_{{H}_{2} O}}{{\left[A\right]}_{C C {l}_{4}}} = \frac{1}{4} = 0.25$

So, you dissolve 1.000 g of $\text{A}$ in 100 mL of water. Now you perform the first extraction by using 100 mL of $C C {l}_{4}$.

Since you're dealing with the ratio between two concentrations of the same substance, you can write

$C = \frac{n}{V} = \text{moles"/"L" = ("g"/"molar mass")/"L" = "g"/"L}$

since the molar mass of $\text{A}$ will cancel out when doing a ratio.

${\left[A\right]}_{C C {l}_{4}} / \left({\left[A\right]}_{{H}_{2} O}\right) = 4$

Assume that x is the mass extracted, which means that the mass remaining in water will be 1.000 - x. The ratio becomes

K_D = (x/cancel("100 mL"))/((1.000-x)/(cancel("100 mL"))) = 4 => x = 4 * 1000 - 4x => x = 0.800

The first extraction will remove 0.800 g of $\text{A}$ from the water. Now for the second extraction. This time, the mass of $\text{A}$ remaining in water will be

${m}_{\text{A" = 1.000 - 0.800 = "0.200 g}}$

The ratio becomes

${K}_{D} = \frac{x}{0.200 - x} = 4 \implies x = 0.800 - 4 x \implies x = \text{0.160 g}$

The mass of "A that remains in water will be

${m}_{\text{A" = 0.200 - 0.160 = "0.040 g}}$

The third extraction will remove

${K}_{D} = \frac{x}{0.040 - x} = 4 \implies x = 0.160 - 4 x \implies x = \text{0.032 g}$

The total mass of $\text{A}$ extracted will be

${m}_{\text{extracted" = 0.800 + 0.160 + 0.032 = "0.992 g}}$

This is equivalent to

"0.992 g" = color(green)(124/125 "g")#