An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following.
Molality of the solution is

Options

A. `0.322`
B. `0.222`
C. `0.413`
D. `0.5`

Answer

A

Explanation

Molarity of solution
`DeltaT_(f)=K_(f) xx m rArr =(DeltaT_(f))/(K_(f))`
`DeltaT_(f)=273-272.4 = 0.6K`,`K_(f)=0.86 K kg mol^(-1)`
`:.m=0.6/1.86=0.322`