Question

An aqueus solution freezes at 272.07 K while pure water freezes at 273 K. Determine the molality and boiling point of the solution. Given `K_(f)=1.86 K//m, K_(b)=0.512 K//m.`

Answer 1

Calcultion of molality of solution
`DeltaT_(f)=K_(f)xxm or m=DeltaT_(f)//K_(f)`
`T_(f)=273-272.07=0.93K, K_(f)=1.86 K kg mol^(-1)`.
`m=((0.93K))/((1.86K kg mol^(-1)))=0.5 molkg^(-1)=0.5m`
Calcultion of boiling point of solution
`DeltaT_(b)=K_(b)xxm,DeltaT_(f)=K_(f)xxm`
`(DeltaT_(b))/T_(f)=(K_(b)xxm)/(K_(f)xxm)=K_(b)/K_(f) or DeltaT_(b)=K_(b)/K_(f)xxDeltaT_(f)`
`DeltaT_(b)=((0.512K kg mol^(-1)))/((1.86K kg mmol^(-1)))=(0.93K)=0.256K`
Boiling point of solution=(373K+0.256K)=373.256 K.