The boiling point of 0.1molal $K_{4} [F e {(C N)}_{6}]$ solution will be : $(G i v e n K_{b} f o r w a t e r = 0.52 K k g {m o l}^{- 1})$

{100.52}^{0} C

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{100.104}^{0} C

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{100.26}^{0} C

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{102.6}^{0} C

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Solution

The correct option is C ${100.26}^{0} C$
Van't hoff factor for $K_{4} [F e (C N)_{6}]$
$= 5$
$Δ T_{b} = i K_{b} m_{s o l u t e}$
$\Rightarrow 5 \times .52 \times .1$
$= 0.26$
$T_{b} = 100 + .26$
$= {100.26}^{0} C$

Option C is correct.

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