Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.22 m SnCl₄(aq). Constants may be found here.

Answer 1

Answer:

T° freezing solution → -11.3°C

T° boiling solution → 103.1 °C

Explanation:

Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"

This salt dissociates as this:

SnCl₄ (aq)  →   1Sn⁴⁺ (aq)  +   4Cl⁻  (aq)   (so i =5)

The formula for the colligative property of freezing point depression and boiling point elevation are:

ΔT = Kf . m . i

where ΔT = T° freezing pure solvent - T° freezing solution

ΔT = Kb . m . i

where ΔT = T° boiling solution - T° boiling pure solvent

Freezing point depression:

0° - T° freezing solution = 1.86°C/m . 1.22 m . 5

T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C

Boiling point elevation:

T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5

T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C