Question

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

Colligative Constants
Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:
Solvent - Formula -
`K_f` value*(°C/m) - Normal freezing point (°C) -
`K_b` value (°C/m) - Normal boiling point (°C)
water - H₂O - 1.86 - 0.00 - 0.512 - 100.00
benzene - C₆H₆ - 5.12 - 5.49 - 2.53 - 80.1
cyclohexane - C₆H₁₂ - 20.8 - 6.59 - 2.92 - 80.7
ethanol - C₂H₆O - 1.99 - -117.3 - 1.22 - 78.4
carbon tetrachloride - CCl₄ - 29.8 - -22.9 - 5.03 - 76.8
camphor - C₁₀H₁₆O - 37.8 - 176
*When using positive
`K_f` values, assume that ?
`T_f` is the absolute value of the change in temperature. If you would prefer to define ?
`T_f` as "final minus initial" temperature, then ?
`T_f` will be negative and so you must use negative
`K_f` values. Either way, the freezing point of the solution should be lower than that of the pure solvent.

`T_f` = _______ Celsius

`T_b` = _______ Celsius

Answer 1

T° freezing solution → -11.3°C

T° boiling solution → 103.1 °C

Step-by-step 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