A solution of a non-volatile solute in water freezes at `-0.40^(@)C`. The vapour pressure of pure water at `298K` is 23.51 torr. For water,`K_(f)=1.86k mol^(-1)kg`.Thus ,vapour pressure of the solution (in torr)

Options

A. `23.44`
B. `19.64`
C. `20.41`
D. `22.47`

Answer

B

Explanation

For `((w_(1))/(m_(1)))` mole of non -volatile solute in `w_(2)`g solvent,
depression in freezing point is
`DeltaT_(f)=(1000K_(f)w_(1))/(m_(1)w_(2))`
`0.40=(1000xx1.86w_(1))/(m_(1)w_(2))`
`:.(w_(1))/(m_(1)w_(2))=(40)/(1000xx186)`
By Raoult's law `(P^(@)-P_(s))/(P^(@))=(n_(1))/(n_(1)+n_(2))~~(n_(1))/(n_(2))=(w_(1))/(w_(2))=(w_(1))/(m_(2))m_(2)/(w_(2))`
`1-(P_(s))/(P^(@))=(40xx18)/(1000xx186)=(72)/(18600)`
`1-(P_(s))/(23.51)=3.87xx10^(-3)`
`P_(s)=19.64` torr