The answer is (4) "124/125 g"
Here's how the problem looks like
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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 Cl_4, and in water.
Since you know that "A" is more soluble in C Cl_4 than it is in water, you can write
K_D = ([A]_(C Cl_4))/([A]_(H_2O)) = 4, or K_D^(-1) = ([A]_(H_2O))/([A]_(C Cl_4)) = 1/4 = 0.25
So, you dissolve 1.000 g of "A" in 100 mL of water. Now you perform the first extraction by using 100 mL of C Cl_4.
Since you're dealing with the ratio between two concentrations of the same substance, you can write
C = n/V = "moles"/"L" = ("g"/"molar mass")/"L" = "g"/"L"
since the molar mass of "A" will cancel out when doing a ratio.
[A]_(C Cl_4)/([A]_(H_2O)) = 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 "A" from the water. Now for the second extraction. This time, the mass of "A" remaining in water will be
m_"A" = 1.000 - 0.800 = "0.200 g"
The ratio becomes
K_D = x/(0.200 - x) = 4 => x = 0.800 - 4x => x = "0.160 g"
The mass of "A that remains in water will be
m_"A" = 0.200 - 0.160 = "0.040 g"
The third extraction will remove
K_D = x/(0.040 - x) = 4 => x = 0.160 - 4x => x = "0.032 g"
The total mass of "A" extracted will be
m_"extracted" = 0.800 + 0.160 + 0.032 = "0.992 g"
This is equivalent to
"0.992 g" = color(green)(124/125 "g")
