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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesMedium2 minPYQ_2017

ChemistryMediumsingle choice

$5 g$ of ${Na}_{2} {SO}_{4}$ was dissolved in $x g$ of $H_{2} O$ . The change in freezing point was found to be $3 {.82}^{o} C$ . If ${Na}_{2} {SO}_{4}$ is $81.5 %$ ionised, the value of $x$

( $K_{f}$ for water $= 1 {.86}^{o} C kg {mol}^{- 1}$ ) is approximately:

(molar mass of $S = 32 g {mol}^{- 1}$ and that of $Na = 23 g {mol}^{- 1}$ )

Options:

Answer:

Solution:

${Na}_{2} {SO}_{4} \to 2 {Na}^{+} + {SO}_{4}^{2 -}$

$i = 1 + (3 - 1) 0 .815 = 2 .63$

$3 .82 = 1 .86 \times 2 .63 \times \frac{5 \times 1000}{142 \times x}$

$∴ x = \frac{1 .86 \times 2 .63 \times 5000}{142 \times 3 .82}$

$= 45 gm$

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

⏱ 2mℹ️ Source: PYQ_2017

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