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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

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The correct Answer is:
B
`Delta T_(b) = K_(b) xx m = 0.52 xx 0.1 = 0.052^(@) C ` `T_(b) = T^(@) + Delta T_(b) = 100^(@) C + 0.052^(@)C=100.052^(@)C`
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