Question

In Southern Ontario, during freezing weather CaCl2 is spread on icy highways to melt the ice. Calculate the freezing point lowering and freezing point of a solution containing 250 g of CaCl2 in 500.0 g of water.

Answer 1

Mass of CaCl₂ = 250 g

Molar mass of CaCl₂ = 110.98 g/mol

Moles of CaCl₂ = mass of CaCl₂/molar mass of CaCl₂

= 250 g/110.98 g/mol

= 2.25 mol

Mass of water = 500.0 g = 0.500 kg ( 1 g = 0.001 kg)

Molality of the solution (m) = moles of solute/kg of solvent

= 2.25 mol/0.500 kg

= 4.50 m

CaCl₂ is a strong electrolyte. In aqueous solution, CaCl₂dissolves to give three particles - one Ca²⁺ and two Cl^-.

Hence, the van't Hoff factor for CaCl₂ (i) = 3

Freezing point depression constant of water (K_f) = 1.86 °C/m

Now, freezing point depression,

or, = 3 x 1.86 °C/m x 4.50 m

or, = 25.1 °C

or, freezing point of water - freezing point of the solution = 25.1 °C

or, 0 °C - freezing point of the solution = 25.1 °C

or, freezing point of the solution = - 25.1 °C

Hence, the freeazing point lowering = 25.1 °C and the freezing point of the solution = - 25.1 °C