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 K_(b)=0.52, the boiling point of this solution will be

Answer»

<P>`102^(@)C`
`103^(@)C`
`101^(@)C`
`100^(@)C`

SOLUTION :`(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`


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