Calculate the freezing point of an aqueous solution having mole fraction of water `0.8`. Latent heat of fusion of ice is `1436. "cal mol"^(-1)`?

Explanation

Given `I_(s)=14363.3 "cal"//"mole" =(1436.3)/(18)="cal"//g`
`K_(t)=(RT^(2))/(1000l_(t))=(2xx273xx273)/(1000xx(1436.3)/(18))`
`K_(t)=(2xx273xx273xx18)/(1000xx1436.3)=1.87`
Now, mole fraction of `H_(2)O=0.8(N)/(n+N)`
`:.` Mole fraction of solute `=0.2=(N)/(n+N)`
`:. (n)/(N)=(0.2)/(0.8)=(1)/(4)`
or `(wxxM)/(mxxW)=(1)/(4)`
or `(wxxM)/(mxxW)=(1)/(18)`
`:.DeltaT=(1000xxK_(1)xxw)/(mxxW)`
`=1000xx.87xx(1)/(4xx18)=25.97^(@)C`
`:."Freezing point" =0-25.97=-25.97^(@)C`