1.

When 0.6 g of urea is dissolved in 100 g of water, the water will boil at ( K_(b) for water = 0.52 Km^(-1) and normal boiling point of water = 100^(@)C ) :

Answer»

372.48 K
373.52 K
373.052 K
273.52 K

Solution :`DeltaT_(B)=(K_(B)xxW_(B)xx1000)/(W_(A)xx M_(B))`
`=(0.52xx0.6xx1000)/(100xx60)=0.052^(@)C`
Boiling point of solution `=373+0.052`
`=373.052`


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