The answer is (4) #"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 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")#
