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")#