Question

A solution made by dissolving 9.81 g of a nonvolatile nonelectrolyte in 90.0 g of water boiled at 100.37degree C at 760 mm Hg. What is the approximate molecular weight of the substance? (For water, Kb = 0.51 degree C/m)

Answer 1

Answer is: molecular weight of the substance is 79.75 g/mol.
Kb(H₂O) = 0,51°C/m.
T(nonvolatile nonelectrolyte) = 100.37 °C.
T(H₂O) = 100°C.
ΔT = 100.37°C - 100°C.
ΔT = 0.37°C.
ΔT = b(nonvolatile nonelectrolyte) · Kb(H₂O).
b(nonvolatile nonelectrolyte) = ΔT ÷ Kb(H₂O).
b(nonvolatile nonelectrolyte) = 0.37°C ÷ 0.51°C/m.
b(nonvolatile nonelectrolyte) = 1.372 mol/kg.
b - molal concentration.
b = n(nonvolatile nonelectrolyte) / m(H₂O).
n = 1.372 mol/kg · 0.09 kg = 0.123 mol.
M(nonvolatile nonelectrolyte) = m / n.
M = 9.81 g ÷ 0.123 mol = 79.75 g/mol.