A solution of urea (mol. Mass 56 g `mol ^(-1)`) boils at `100.18^(@)C` at the atmospheric pressure. If `K_(f) and K_(b)` for water are `1.86 and 0.512 K kg mol ^(-1)` respectively, the above solution will freeze at

Options

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

Answer

B

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))`
`implies [Delta T _(b) =100. 18-100= 0.18^(@)C]`
`= 0.18xx (1.86)/(0.512) =0.654^(@)C`
As the Frecezing Point of pure water is `0^(@)C, Delta T _(f) =0-T_(f)`
`0.654 =-0T_(f)`
`therefore T _(f) =-0.654`
Thus the freezing point of solution will be `- 0.654 ^(@)C.`