Calculate the freezing point of a solution of a non-volatile solute in a unknown solvent of molar mass `30g//"mole"` having mole fraction of solvent equal to `0.8`. Given that latent heat of fusion of solid solvent `=2.7 "kcal mol"^(-1)`, freezing point of solvent `=27^(@)C` and `R=2 "cl mol"^(-1)K^(-1)`.

Answer

`10.33^(@)C`

Explanation

`because K_(f)=(MRT_(f)^(2))/(1000DeltaH_(f))=(2xx300xx300xx30)/(1000xx2700)=2.00K kg "mol"^(1)`
`because` mole fraction of waer `=0.8`
`therefore` mole fraction of solute `=0.2`
or `0.2=(n)/(n+N)` and `0.8=(N)/(n+N)`
`therefore (n)/(N)=(0.2)/(0.8)=(1)/(4)=(wM)/(mW)`
or `(1)/(4)=(wxx30)/(mW) therefore (w)/(mW)=(1)/(4xx30)`
`therefore DeltaT_(f)=(100xxwxxK_(f))/(mxxW)=(1000xx2.0)/(4xx30)=16.67`
`therefore F.P.` of solution. `=10.33^(@)C`.