A6.0 gram sample of an unknown compound is dissolved in 125 g of water. the solutions freezing point is lowered to 1.38 c° below the freezing point of pure water. what is the molar mass of the sample? k for water is 1.86 c°/m.

Change in the freezing point = Kf * m

m = molality
Kf = 1.86 °C/m

m = Change in the freezing point / k = 1.38°C / 1.86 °C/m =0.742 m

m = #mol of solute / Kg of solvent

Kg of solvent = 125g/1000 g/kg = 0.125 kg.

#mol of solute = m* Kg of solvent = 0.742m * 0.125 kg =0.09275mol

Molar mass = 6. 0 g / 0.09275 mol = 64.7 g/mol

The density would be 1.622