Answer to Question #312673 in General Chemistry for Sam

Question

What is the mole fraction of the solute in a 40% by mass ethanol(C2H6O) solution in water?

Answer 1

X(C₂H₆O) = 40 %;

M(C₂H₆O) = 46 g/mole;

M(H₂O) = 18 g/mole;

If m(solution) = 100 g;

Then m(C₂H₆O) = 40 g and m(H₂O) = 60 g.

n(C₂H₆O) = m(C₂H₆O)/M(C₂H₆O) = 40/46 = 0.87 moles;

n(H₂O) = m(H₂O)/M(H₂O) = 60/18 = 3.33 moles;

n(total) = n(C₂H₆O) + n(H₂O) = 0.87 + 3.33 = 4.20 moles;

x(C₂H₆O) = n(C₂H₆O) * 100%/n(total) = 0.87 * 100/4.20 = 20.7 %.

Answer: 20.7 %.