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^@C`
B. `-0.654^@C`
C. `6.54^@C`
D. `-6.54^@C`

Answer

B

Explanation

As `DeltaT_f = K_f.m`
`Delta T_b = K_b.m`
Hence,we have `m = (DeltaT_f)/(K_f) = (DeltaT_b)/K_b`
or `DeltaT_f = DeltaT_b (K_f)/(K_b) rArr [DeltaT_b = 100.18 - 100 = 0.18^@C]`
`:.DeltaT_f = 0.18 xx 1.86/0.512 = 0.654^@C`
As the freezing point of pure water is `0^@C`,
`DeltaT_f = 0-T_f`
`0.654 = 0 -T_f`
`:. T_f = -0.654`
Thus the freezing pointof solution will be `-0.654^@C`.