Question

What is the freezing point and boiling point respectively of an aqueous solution of LiCl that contains 2.58 g of LiCl and 10.0 mL of water?

Answer 1

Answer:

М(LiCl) = (6.94+35.5) = 42.4 g/mol

n(LiCl) = 2.58g / 42.4g/mol = 6.08 x 10-2 mol

This amount of LiCl with dissociate to produce 1.22 x 10-1 mol of ions

10.0ml H2O * (1g/ml) * (1Kg/1000g) = 1.00 x 10-2 Kg solvent

molality of solute = 1.22 x 10-1 mol/ 1.00 x 10-2 Kg ) = 12.2 m

DTf = -Kf * m = -1.86 °C m-1 * 12.2m = -22.7 °C

Freezing point = 0°C - 22.7°C = -22.7°C

DTb = Kb * m = 0.512°C m-1 * 12.2m = 6.25 °C

Boiling point = 100°C + 6.25°C = 106 °C