Answer is: the freezing point of the solution of sucrose is -0.435°C.
m(H₂O) = 100 g ÷ 1000 g/kg = 0.1 kg.
m(C₁₂H₂₂O₁₁) = 8.0 g.
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 8.0 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0233 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0233 mol ÷ 0.1 kg.
b(solution) = 0.233 m.
ΔT = b(solution) · Kf(H₂O).
ΔT = 0.233 m · 1.86°C/m.
ΔT = 0.435°C.
Tb = 0°C - 0.435°C = -0.435°C.
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