Assuming 100% dissociation, calculate the freezing point ( 𝑇f ) and boiling point ( 𝑇b ) of 2.53 𝑚 CaCl2(aq) .

Answer 1

∆T = imK

∆T = change in boiling point = ?

i = van't Hoff factor = 3 for CaCl₂ since it dissociates into 1Ca²⁺ and 2Cl^- (3 particles)

m = molality = moles / kg solvent = 2.53 m

K = boiling point constant for water = 0.512º/m or freezing point constant for water = 1.86º/m

Dissolved solutes LOWER the freezing point and RAISE the boiling point

Freezing point elevation:

∆T = (3)(2.53)(1.86)

∆T = 14.1º

Freezing point of the solution = 0º - 14.1º = -14.1ºC (3 sig. figs.)

Boiling point elevation:

∆T = (3)(2.53)(0.512)

∆T = 3.89º

Boiling point of solution = 100º + 3.89º = 104ºC (3 sig. figs.)