Question

Determine the ideal value of the van’t hoff factor for Al₂(SO₄)₃ at infinite dilution. Then, use that value to determine the freezing point of a solution when 15.00 g is dissolved into 150 mL of water. Round your answer to the tenths digit

Answer 1

1 molecule of Al2(SO4)3 will break into 2 ions of Al⁺³ and 3 ions of SO4^2- . So ideal Van't Hoff factor(i) is 2+3 = 5.

T = i * Kf * molality of solute

T is depression in freezing point of water.

Kf for water is 1.86 C/m

and molality of solution = Number of moles of solute/ Mass of solvent

So, number of moles of Al2(SO4)3 = 15/MW of Al2(SO4)3 = 15/ 342.15 = 0.0438

Mass of solvent = 150g = 0.15 kg ( Density of water = 1g/ml)

So, molality = 0.0438/0.15

= 0.292m

SO,

T = 5 * 1.86 C/m * 0.292m

= 2.718 C

So, freezing point of solution = Actual freezing point of water - T

= 0C - 2.718C

= -2.718C or 270.28K