Answer to Question #177674 in Chemistry for amelia

Question

Camphor (C10H16O) melts at 179.8°C and has a large freezing point depression constant, Kf = 40.00°C. m–1 . When 186.0 mg of an unknown nonvolatile, non-electrolyte compound is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture was found to be 176.7°C. What is the molar mass of the unknown compound?

Answer 1

Solution:

The freezing point depression (ΔT_f) is related to the molality (m) and the freezing point depression constant (K_f) by ΔT_f = K_f × m (for non-electrolyte compounds).

ΔT_f = 179.8°C - 176.7°C = 3.1°C

Hence,

m = ΔT_f / K_f = (3.1°C) / (40.00°C m^–1) = 0.0775 m = 0.0775 mol/kg

m = 0.0775 mol/kg

Molality (m) = Moles of solute / Mass of solvent (kg),

Hence,

Moles of solute = Molality × Mass of solvent (kg) = 0.0775 mol/kg × 0.02201 kg = 0.001706 mol

Mass of solute = Moles of solute × Molar mass of solute

Hence,

Molar mass of solute = Mass of solute / Moles of solute

Molar mass of solute = (0.186 g) / (0.001706 mol) = 109.027 g/mol = 109 g/mol

Molar mass of solute = 109 g/mol

Answer: The molar mass of the unknown compound is 109 g/mol.