The freezing point of a solution made by dissolving 12.4 g nonelectrolyte in 100.0 g water is -5.00celcuis. calculate the molar mass of the electrolyte

46.13 g/mol.

Explanation:

∵ ΔTf = Kf.m.

Where,  ΔTf is the depression in the freezing point of water = 5.0°C.

Kf is the the molal freezing point depression constant of water = 1.86 °C/m.

m is the molality of the solution (moles of solute per kilogram of solvent.)

∵ ΔTf = Kf.m.

∴ m = ΔTf /Kf = (5.0°C)/(1.86 °C/m) = 2.688 m.

∵ m = no. of moles of solute / kg of solvent.

∴ 2.688 m = (mass of solute / molar mass of solute) / kg of water.

∴ Molar mass of solute = (mass of solute)/(2.688)(kg of water) = (12.4 g)/(2.688)(0.10 kg) = 46.13 g/mol.

- lead + phosphoric acid = hydrogen + lead(ii) phosphate