1.

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

Answer»

`101^(@)C`
`100^(@)C`
`102^(@)C`
`103^(@)C`

Solution :`((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`


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