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

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

A

`373.052 K`

B

`273.52K`

C

`372.48K`

D

`273.052K`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaT_(b) = ((0.6)/(60))/(100) xx 1000 xx 0.52`
`= (1)/(1000) xx 1000 xx 0.52 = 0.52`
`T - 373 = 0.52 T = 373.52`
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