Calculate the freezing point of a solution of urea in water which boils at `100.18^@C`.
`(K_f = 1.86 ^@C m^(-1) K_b = 0.512^@C m^(1))`

Options

A. `- 0.18^@C`
B. `0.65^@C`
C. `- 0.65^@C`
D. `0.18^@C`

Answer

C

Explanation

`m= (DeltaT_f)/(K_f) = (DeltaT_b)/(K_b) (DeltaT_b =100.18 – 100 = 0.18 ^@C)`
`DeltaT_f = (0.18xx1.86)/(0.512) = 0.65^@C`
`DeltaT_f = T_f^@ implies T_f = 0 - 0.65 = - 0.65^@C` (freezing point of pure water = `0^@C`).