The vapour pressur of `0.01molal` solution of weak base `BOH` in water at `20^(@)C` is `17.536mm`. Calculate `K_(b)` for base. Aqueous tension at `20^(@)C` is `17.540mm`. Assume molatilly and molarity same.

Explanation

We have from Raoult's law
`(p^(0)-P_(s))/(p^(0)) = x_(2) = (17.54-17.536)/(17.54) = 0.00022`
`m = (x_(2)xx1000)/(x_(1)xxMw_(1)) = (0.00022 xx 1000)/(0.999 xx 10) = 1.267 xx 10^(-2)`
Molarity `=` Concentration of `[BOH] 1.267 xx 10^(-2)`
For `{:(BOHhArr,B^(o+)+,overset(Θ)OH,,),(1,0,0,,),((1-alpha),alpha,alpha,,):}`
Molality is also given as `1 xx 10^(-2)`.
`:. ("Exp. value of molality")/("Normal value of molality") = 1+ alpha`
`:. (1.267 xx 10^(-2))/(1xx10^(-2)) = 1+alpha :. alpha = 0.267`
Now
`K_(b) = (Calpha^(2))/((1-alpha)) = (0.01 xx 0.267xx0.267)/((1-0.267)) = 9.74 xx 10^(-4)`