Answer to Question #323407 in General Chemistry for Ashter

Question

Camphor (C10H16O2) freezes at 175 °C, and it has a particularly large freezingpoint-depression constant, Kf = 40.0 °C/m. When 0.086 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 1.7 °C below of pure camphor. What is the molar mass of the solute?

You add 0.25 g of an unknown solute to 11.1 g of benzene. The boiling point of the benzene rises from 80.10 °C to 80.46 °C. What is the molar mass of the compound? Kb=2.53 °C /m

A solution of an unknown non-dissociating solute was prepared by dissolving 0.300 g of the substance in 40.0 g of CCl4. The boiling point of the solution was 0.357 °C higher than that of the pure solvent. Calculate the molar mass of the solute. Kb=5.02 °C/m

Answer 1

1) ∆T = K_f × m

1.7 °C = 40.0 °C/m × m

m = 1.7 / 40.0 = 0.0425 m

m = n(solute) / kg (solvent) = m(solute) / (Mr(solute) × kg(solvent))

0.0425 m = (0.086 g × 1000) / (Mr(solute) × 22.01 g

Mr(solute) = 91.93 g/mol

2) ∆T = K_b × m

∆T = T_solution - T_pure = 80.46°C - 80.10°C = 0.36°C

0.36 °C = 2.53 °C/m × m

m = 0.36 / 2.53 = 0.142 m

m = n(solute) / kg (solvent) = m(solute) / (Mr(solute) × kg(solvent))

0.142 m = (0.25 g × 1000) / (Mr(solute) × 11.1 g

Mr(solute) = 158.6 g/mol

3) ∆T = K_b × m

0.357 °C = 5.02 °C/m × m

m = 0.357 / 5.02 = 0.071 m

m = n(solute) / kg (solvent) = m(solute) / (Mr(solute) × kg(solvent))

0.071 m = (0.300 g × 1000) / (Mr(solute) × 40.0 g

Mr(solute) = 105.6 g/mol