Question

A solution is made by dissolving 10.2 grams of glucose (c6h12o6) in 0.315 kilograms of water. if the molal freezing point constant for water (kf) is -1.86 °c/m, what is the resulting δtf of the solution? show all of the steps taken to solve this problem.

Answer 1

Answer is: resulting ΔTf of the solution is 0.334°C.
m(glucose-C₆H₁₂O₆) = 10.2 g.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 10.2 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆) = 0.057 mol.
b(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ m(H₂O).
b(C₆H₁₂O₆) = 0.057 mol ÷ 0.315 kg.
b(C₆H₁₂O₆) = 0.18 mol/kg.
ΔT = Kf · b(C₆H₁₂O₆).
ΔT = 1.86°C/m · 0.18 m.
ΔT = 0.334°C.