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When 0.6 g of urea is dissolved in 100 g...

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` ) :

A

372.48 K

B

373.52 K

C

373.052 K

D

273.52 K

Text Solution

Verified by Experts

The correct Answer is:
C
`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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