A solution of urea boils at `100.18^(@)C` at the atmospheric pressure. If `K_(f)` and `K_(b)` for water are `1.86` and `0.512K kg mol^(-1)` respectively, the above solution will freeze at,

Options

A. `0.654^(@)`
B. `-0.645^(@)C`
C. `6.54^(@)C`
D. `-6.54^(@)C`

Explanation

As ` Delta T _ f = K _ f m `
` Delta T _ b = K _ b m`
Hence, we have ` m = ( Delta T _ f ) / ( K _ f ) = ( Delta T _ b ) /( K _ b ) `
or ` Delta T _ f = Delta T _ b ( K _ f ) /( K _ b ) `
` rArr [ Delta T _ b = 100.18 - 100 = 0.18 ^@C] `
` = 0.18 xx (1.86 ) / (0.512 ) = 0.654 ^@ C `
As the Freezing Point of pure water is ` 0 ^@ C`,
` Delta T _ f = 0 - T _ f `
` 0.654 = 0 - T _ f `
` therefore T _ f = - 0.654 `
Thus the freezing point of solution will be ` - 0.654 ^@ C`.