`K_(f)` for water is `1.86^(@)C*m^(-1)`. What is the molality of a solutionn which freezes at `-0.192^(@)C`? Assuming no change in the solute by raising the temperature, at what temperature will the solution boil ? (`K_(b)` for `H_(2)O=0.515^(@)C*m^(-1)`)

Explanation

Molality `=(DeltaT_(f))/(K_(f))`
`=(0.192)/(1.86)=0.103m`
Again `DeltaT_(b)=K_(b).m` …………(Eqn. 8)
`=0.515xx0.103=0.0532^(@)C`
Assuming the b.p. of pure water to be `100^(@)C` the b.p. of the solution will be `100.0532^(@)C`