Question

A solution is prepared by dissolving 38.7g of nonelectrolyte into 218g of water.

The freezing point of the solution is measured to be -5.53°C. Calculate the

molar mass of the solute. kf of water = 1.86°C/m

Answer 1

Freezing point depression= Freezing point of water-Freezing point of solution

= 0-(-5.53)= 5.53 °C

Freezing point depression= Freezing point constant of water $\times$ molality

molality = Freezing point depression/Freezing point constant of water

molality $= \frac{5.53}{1.86}= 2.973 \; m$

molaity=moles of solute/mass of water in kg

moles of solute= molality $\times$ mass of water in kg

moles of solute $= 2.973 \times 0.218= 0.648$ moles of solute

molar mass of solute $= \frac{mass}{moles} = \frac{38.7}{0.648}= 59.7 \;g/mol = 60 \; g/mol$