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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesMedium2 minQB

ChemistryMediumsingle choice

The van't Hoff's factor of $K_{4} [Fe (CN)_{6}]$ assuming $50%$ dissociation is:

Answer:

Solution:

$K_{4} [Fe (CN)_{6}] \to 4 K^{+} + [F e (CN)_{6}]^{4 -}$

Therefore,

$i = 1 + (n - 1) α$

$i = 1 + (5 - 1) \times 0.5$

$i = 3$

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Abnormal Colligatiove Properties

⏱ 2mℹ️ Source: QB

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