(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol Let a be the number of millimoles of I2 extracted Millimoles of I2 left in aqueous phase = 0.5 - a Distribution coefficient D = [I2](organic)/[I2](aqueous) = [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)] = [(a/50)]/[(0.5 - a)/50] = 85 a/(0.5 - a) = 85 a = 0.4942 mmol Thus amount of I2 extracted = 0.4942 mmol (b) Amount of I2 left in aqueous phase = 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol Concentration of I2 in aqueous phase = moles/volume (in L) = 5.8 x 10-6/0.050 = 1.16 x 10-4 M Solution (a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol Let a be the number of millimoles of I2 extracted Millimoles of I2 left in aqueous phase = 0.5 - a Distribution coefficient D = [I2](organic)/[I2](aqueous) = [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)] = [(a/50)]/[(0.5 - a)/50] = 85 a/(0.5 - a) = 85 a = 0.4942 mmol Thus amount of I2 extracted = 0.4942 mmol (b) Amount of I2 left in aqueous phase = 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol Concentration of I2 in aqueous phase = moles/volume (in L) = 5.8 x 10-6/0.050 = 1.16 x 10-4 M.