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

2

Explanation

Boiling point of the solution= boiling point of solvent `(T_(b))`+ elevation of boiling point `(DeltaT_(b))`
`DeltaT_(b)=I K_(b)m`
`=I K_(b)(w_(solute)//M_(solute))/w_(solvent)xx(1000 g)/(kg)`
Since chemical identify of solute is not mentioned, `i=1`. To calculate to `M_(solute)`, we use second Raoult's law:
`(P_(1)^(0)-P)/P_(1)^(0) n_(solute)/(n_(solute)+n_(solvent))`
`Al 100^(@)C` the bpt of water, the `P_(1)^(0)=760 mm Hg`. Thus
`((760-732) mm Hg)/(760 mm Hg)=(65g//M_(solute))/(100 g..18 g mol^(-1))`
`M_(solute)=((6.5 g)(18 g mol^(-1))(760))/((100 g)(28))`
`=31.75 g mol^(-1)`
Substituting this results we have
`DeltaT_(b)=((0.52 k kg mol^(-1))(6.5 g)(1000 g kg^(-1)))/((31.75 g mol^(-1))(100 g))`
`=1.06 K` or `1.06^(@)C`
Thus, boiling point of solution
`=100^(@)C+1.06^(@)C`
`=101.06^(@)C`