SOLUTION CHEMISTRY
Determine the boiling point and freezing point of a solution of 0.30 g of glycerol (C3H8O3) in 20.0 g of water.
Freezing point = 0.00 oC + .30 oC = -.30 oC
DTf = (-1.858 oC/m)( .16m) = -.30 oC
Boiling point = 100.00 + .083 = 100.083 oC
DTb = (.521 oC/m)( .16 m) = .083 oC
molality of solution = .0033 moles / .020 kg = .16 m
moles glycerol = (.30 g) (1 mole / 92 g) = .0033 moles