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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2021

ChemistryEasynumerical

When $12.2 g$ of benzoic acid is dissolved in $100 g$ of water, the freezing point of solution was found to be $- 0.93 ° C (K_{f} (H_{2} O) = 1.86 K kg {mol}^{- 1})$ . The number $(n)$ of benzoic acid molecules associated (assuming $100 %$ association) is _.

Answer:

2.00

Solution:

${ΔT}_{f} = i \times k_{f} \times m$

$0 - (- 0.93) = i \times 1.86 \times \frac{12.2}{122 \times 100} \times 1000$

$i = \frac{0.93}{1.86} = 0.5$

$i = 1 + (\frac{1}{n} - 1) α$

$\frac{1}{2} = 1 + (\frac{1}{n} - 1) \times 1$

$n = 2$

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

⏱ 2mℹ️ Source: PYQ_2021

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