If boiling point of an aqueous solution is `100.1^(@)C`. What is its freezing point? Given latent heat of fusion and vaporization of water are `80calg^(-1)` and `540 calg^(-1)` respectively.

Explanation

`DeltaT_(b)=K_(b)'xx"molality"`
`DeltaT_(t)=K_(t)'xx"molality"`
`therefore (Delta_(b))/(DeltaT_(t))=(K'_(b))/(K_(f))=(RT_(b)^(2))/(1000|v)xx(1000|f)/(RT_(f)^(2))`.
`(DeltaT_(b))/(DeltaT_(f))=(K_(b)^(2)xx|_(f))/(T_(f)^(2)xx|_(v))`
`T_(b)=100+273==373K`.
`T_(t)=0+273=273K`.
`l_(t)=80 "cal" g^(-1)`.
`l_(t)=540 "cal" g^(-1)`.
`therefore (0.1)/(DeltaT_(f))=(373xx373xx80)/(273xx273xx540)`.
`therefore DeltaT_(f)=0.362`.
`therefore T_(t)=0.0-0.362.=-0.362^(@)C`.