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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2020

ChemistryEasysingle choice

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. The apparent degree of dissociation of $N a_{2} S O_{4}$ is

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Answer:

Solution:

$P (N a_{2} S O_{4}) = i C R T = i (0.004) R T$
P (glucose) $=$ CRT $=$ 0.010 RT
As solutions are isotonic, i (0.004) RT $=$ 0.01 RT
this gives i $=$ 2.5
$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
$α = \frac{i - 1}{n - 1}$
Here $n = 3$
$α = \frac{2.5 - 1}{2} = \frac{1.5}{2} = 0.75 = 75 %$

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

⏱ 2mℹ️ Source: PYQ_2020

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