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

Answer

B

Explanation

At B.P. `P_(0)=760` torr
`(P^(0)-P_(s))/(P_(s))=(W_(A)//M_(A))/(W_(B)//M_(B))`
`(760-732)/(732)=(6.5//M)/(100//18)`
On solving M = 32.
for elevation of B.P
`? T_(B)=I K_(b)m=1xx0.52(((6.5)/(32)))/(100)xx1000=1`
So `B.P=100+ ? T_(B)=101^(@)C`