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_(2))/(n_(1))=(n_(2))/(w_(1)//M_(1))`
`or n_(2)=(28)/(760)xx(100)/(18)="0.2046 mole"`
`"Molality of the solution"=("0.2046 mole")/("100/1000 kg")`
`="2.046 mole kg"^(-1)`
`DeltaT_(b)=K_(b)xx"molality"=0.52xx2.-46=1.06^(@)`
i.e., `T_(b)-T_(b)^(@)=1.06^(@)`
`or T_(b)=T_(b)^(@)+1.06=100+1.06^(@)C=101.06^(@)C`