Chemistry help please?

Question

Iodine dissolves in water, but its solubility in a nonpolar solvent such as CCl4 is greater.

The equilibrium constant is 85.0 for the reaction

I2 (aq) <---> I2 (CCl4)
 
You place 0.0340 g of I2 in 100.0 mL of water. After shaking it with 10.0 mL of CCl4, how much I2 remains in the water layer?

I have been told the answer is  2E-5 mol which is 5x-4 grams

This isn't homework, just studying, but I have been trying to figure out the process and cant. Thanks.

? · Accepted Answer

? MI2 = 0.0340 g I2 / 254 g /mol I2 / 0/1L = 1.34x10E-4 M
Kd = 85 = [I2]org/[I2]aq, let x = M I2 in CCl4, then [1.33x10E-3 - x ] is conc. left in water.
85 = x /(1.33x10E-1 - x)
x = 0.000132 M
? g = 0.000132 X 254 = 0.0335 g in CCl4 phase
g H2O = 0.0340 g - 0.0335g = 0.0005 g = 5x10E-4

Chemistry help please?

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