At `100^(@)C` the vapour pressure of a solution of `6.5g` of an solute in `100g` water is `732mm`.If `K_(b)=0.52`, the boiling point of this solution will be :

Options

A. `102^(@)C`
B. `103^(@)C`
C. `101^(@)C`
D. `100^(@)C`

Answer

C

Explanation

`(p^(@) -p_(s))/(p^(@)) = (n_(B))/(n_(B)) = (w_(B) xx M_(A))/(M_(B) xx w_(A))`
`(760-732)/(760) = (6.5 xx 18)/(M_(B) xx 100)`
`(28)/(760) = (6.5 xx 18)/(M_(B) xx 100)`
`M_(B) = (6.5 xx 18 xx 760 )/(28 xx 100) = 31.76`
Now,`Delta t_(B) = (K_(b) xx w_(B) xx 1000)/(w_(A) xx M_(B)) = (0.52 xx 6.5 xx 1000)/(100 xx31.76) = 1.06`
`therefore ` Boiling point of solution `= 100+1.06 = 101.06^(@)C`
`~~101^(@)C`