A commercial lubricant, Dubdeefordy, has K = SN/SM = 13.1 for immiscible solvent M and N....
A commercial lubricant, Dubdeefordy, has K = SN/SM = 13.1 for immiscible solvent M and N. If 5.7 grams of Dubdeefordy is shaken with 115 mL of solvent M and 45 mL of solvent N, what weight of Dubdeefordy will be found in each layer at equilibrium? Show your calculations.
Homework Answers
Total mass of Dubdeefordy = 5.7 g
Let mass of Dubdeefordy in solvent M = x
then, mass of Dubdeefordy in solvent N = 5.7 g - x
Concentration of Dubdeefordy in solvent M = (mass of Dubdeefordy in solvent M) / (volume of solvent M)
Concentration of Dubdeefordy in solvent M = (x) / (115 mL)
Concentration of Dubdeefordy in solvent N = (mass of Dubdeefordy in solvent N) / (volume of solvent N)
Concentration of Dubdeefordy in solvent N = (5.7 g - x) / 45 mL
K = SN / SM
K = (Concentration of Dubdeefordy in solvent N) / (Concentration of Dubdeefordy in solvent M)
13.1 = [(5.7 g - x) / 45 mL] / [(x) / (115 mL)]
13.1 * (45 mL / 115 mL) = (5.7 g - x) / x
5.126 = (5.7 g - x) / x
5.126x = 5.7 g - x
5.126x + x = 5.7 g
6.126x = 5.7 g
x = (5.7 g) / 6.126
x = 0.93 g
Mass of Dubdeefordy in solvent M = x
Mass of Dubdeefordy in solvent M = 0.93 g
Mass of Dubdeefordy in solvent N = 5.7 g - x
Mass of Dubdeefordy in solvent N = 5.7 g - 0.93 g
Mass of Dubdeefordy in solvent N = 4.77 g
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