Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Chemistry help please?
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.
1 Answer
- HalchemistLv 79 years agoFavorite 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
