Answer to Question #328629 in General Chemistry for Max

Question

Calculate the freezing point of a solution made from 32.7 g of propane, C3H8, dissolved in 137.0 g of benzene, C6H6. The freezing point of benzene is 5.50°C and its Kf is 5.12°C/m.

Answer 1

ΔT_F = K_F · b

ΔT_F - the freezing-point depression, is defined as T_{F (pure solvent)} − T_{F (solution)}

K_F - the cryoscopic constant

b - the molality

b = n_solute /  m_solvent

b = n(C₃H₈) / m(C₆H₆) = (m(C₃H₈)/Mr(C₃H₈)) / m(C₆H₆) = (32.7 g / 44 g/mol) / 0.137 kg = 5.42 mol/kg

ΔT = 5.12°C/m * 5.42 mol/kg = 27.75°C

T_{F (solution) =} ΔT_F - T_{F (pure solvent)}

T_f(C₃H₈) = ΔT - T_f(C₆H₆) = 27.75°C - 5.50°C = 22.25°C