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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2020

ChemistryEasysingle choice

What will be the temperature at which a solution containing 6 g of glucose per 1000 g water will boil if molal elevation constant for water is 0.52 K kg mol^-1.

Options:

Answer:

Solution:

$w = 6 g, W = 1000 g,$ Mol. wt. of glucose = 180
$Δ T_{b} = \frac{1000 \times k_{b} \times w}{m \times W}$
$= \frac{1000 \times 0.52 \times 6}{180 \times 1000} = 0.0173 ° C$
Hence boiling point of solution = b.p. of water + $Δ T_{b} = 100 + 0.0173$
$= 100.0173 ° C$

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

⏱ 2mℹ️ Source: PYQ_2020

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