The boiling point of 0.1 molal $K_{4} [F e (C N)_{6}]$ solution will be ( $K_{b}$ for water = 0.52 K.Kg $m o l e^{- 1}$ )

273^{\circ} C

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{100.26}^{\circ} C

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273 K

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500 K

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Solution

The correct option is B ${100.26}^{\circ} C$
$Δ T_{b} = K_{b} . m . i = 0.52 \times 0.1 \times 5 = {0.26}^{\circ} C$
Boiling point = ${100.26}^{\circ} C$

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