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ChemistryLiquid SolutionMiscellaneous/MixedEasy2 minPYQ_2020

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If a 0.004 M solution of $N a_{2} S O_{4}$ is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of disociation of $N a_{2} S O_{4}$ is

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Answer:

Solution:

$\underset{(0.004 - x)}{N a_{2} S O_{4}} ⇌ \underset{2 x}{2 N a^{+}} + \underset{x}{S O_{4}^{2 -}}$

Since both the solution are isotonic $0.004 + 2 x = 0.01$

$x = 0.003$

$H N O_{2}$ Percent dissociation. $= \frac{3 \times 1 0^{- 3}}{0.004} \times 100 = 75 %$ .

Stream:NTA_ABHYASSubject:ChemistryTopic:Liquid SolutionSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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