Chemistry - Eudiometry Question with Solution | TestHub
ChemistryEudiometryEudiometric CalculationsMedium2 minQB
ChemistryMediumnumerical
A 40 mL gaseous mixture of CO,CH₄,Ne is exploded with 50 mL air (80% N₂,20% O₂ by volume). KOH treatment reduces volume by 8 mL; alkaline pyrogallol reduces it by another 2 mL. Find the percentage CO in the original mixture.
Answer:
13.33
Solution:
Air supplies 0.20×50=10 mL O₂. Pyrogallol removes 2 mL remaining O₂, so 8 mL was consumed. KOH removes 8 mL CO₂. Let CO=x, CH₄=y mL. Then x+y=8 and x/2+2y=8. Solve y=8/3, x=16/3; CO percentage=(16/3)/40×100=13.33%.
Stream:JEESubject:ChemistryTopic:EudiometrySubtopic:Eudiometric Calculations
⏱ 2mℹ️ Source: QB
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