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ChemistryLiquid SolutionAbnormal Colligatiove PropertiesEasy2 minPYQ_2023

ChemistryEasysingle choice

Calculate osmotic pressure of solution of 0.025 mole glucose in $100 mL$ water at $300 K$ . $[R = 0.082 atm dm^{3} mol^{- 1} K^{- 1}]$

Answer:

Solution:

$π = MRT = \frac{n _{2} RT}{V} = \frac{0.025 mol \times 0.082 dm ^{3} atm mol ^{- 1} K ^{- 1} \times 300 K}{0.1 dm ^{3}} = 6.15 atm$

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

⏱ 2mℹ️ Source: PYQ_2023

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