Pamela L. answered • 10/21/17

Patient tutor for organic and general chemistry (all levels)

This set up involved a little bit of algebra. We know some of the sample with stay in the water and some will go into the CH

_{2}Cl_{2}layer. Based upon your value for K, it is more soluble in CH_{2}Cl_{2}so we would expect a higher amount in that layer at the end of our calculation.Let's define x as the mass in the H

_{2}O layer. And then 0.500-x will be the mass in the CH_{2}Cl_{2}layer K = concentration in CH

_{2}Cl_{2}= 5.05 = (0.5-x)/20mL = (0.5-x)(125mL) = 5.05 concentration in H

_{2}O x/125 mL 20x Solve for x = 0.28 g in H

_{2}O And 0.5g -0.28 g = g in CH_{2}Cl_{2}