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

Answer

A

Explanation

`((P^(@)-P_(s))/(P^(@)))=(n)/(N)=(w_("solute"))/(M_("solute"))xx(M_("solvent"))/(W_("solvent"))`
at `100^(@)C, P^(@)=760 mm`
`(760-732)/(760)=(6.5xx18)/(M_("solute")xx100)`
`M_("solute")=31.75 g mol^(-1)`
`Delta T_(b)=m xx K_(b)=(w_("solute")xx1000)/(M_("solute")xxw_("solvent"))xx K_(b)`
`Delta T_(b)=(0.52xx6.5xx1000)/(31.75xx100)=1.06^(@)C`
`therefore` boiling point of solution `= 100^(@)C+1.06^(@)C ~= 101^(@)C`