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K(b) for water is 0.52 K/m. Then, 0.1 m ...

`K_(b)` for water is 0.52 K/m. Then, 0.1 m solution of NaCl will boil approximately at

A

`100.52^(@)C`

B

`100.052^(@)C`

C

`101.04^(@)C`

D

`100.104^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
d

`DeltaT_(b) = iK_(b).m = 2xx0.52xx0.1 - 0.104^(@)C`
`T_(b) = 100 + 0.104^(@)C`
`= 100.104^(@)C`
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