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The boiling point of 0.1 mK(4)[Fe(CN)6] ...

The boiling point of `0.1 mK_(4)[Fe(CN)_6]` is expected to be `(K_(b)` for water = `0.52K kg mol^(-1)`)

A

`100.52^(@)C`

B

`100.10^(@)C`

C

`100.26^(@)C`

D

`102.6^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
`i=1+(n-1)alpha=1+(5-1)xx1=5`
`DeltaT_(b)=iK_(b)xxm=5xx0.52xx1=0.26K`
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