Question

What is the freezing point of an aqueous solution that boils at 106.6 ∘C?

Express your answer using two significant figures.

Answer 1

Answer: The freezing point of solution is
`-24^0C`

Explanation:

Formula used for elevation in boiling point is,

`\Delta T_b=k_b* m`

where,

`T_b` = change in boiling point

`k_b` = boiling point constant=
`0.512^0Ckg/mol`

m = molality

Given: Boiling point of solution =
`106.6^0C`

Boiling point of water=
`100^0C`

thus
`6.6^0C=0.512^0Ckg/mol* m`

`m=12.8`

Formula used for lowering in freezing point is,

`\Delta T_f=k_f* m`

where,

`T_f` = change in freezing point

`k_f` = freezing point constant =
`1.86^0Ckg/mol`

m = molality

Freezing point of water=
`0^0C`

`0^0C-T_f=1.86* 12.8`

`T_f=-24^0C`