What is the freezing point of an aqueous solution that boils at 105.0 ∘c? express your answer using two significant figures.

T°fussion of solution is -18°C

Explanation:

We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1

First of all, we apply boiling point elevation

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb =  ebuliloscopic constant

105°C - 100° = 0.512 °C kg/mol  . m . 1

5°C / 0.512 °C mol/kg = m

9.7 mol/kg = m

Now that we have the molality we can apply, the Freezing point depression.

ΔT = Kf . m . i

Kf =  cryoscopic constant

0° - (T°fussion of solution) = 1.86 °C/m  . 9.76 m . 1

- (1.86°C /m . 9.7 m) = T°fussion of solution

- 18°C = T°fussion of solution

the dissolution of polymer can be thermodynamically described in 2 steps.