Chemistry - Redox Reactions Question with Solution | TestHub
ChemistryRedox ReactionsDouble IndicatorHard2 minQB
ChemistryHardnumerical
A quantity of 0.88 g was burnt in excess of air and the resultant gases ( and ) were passed through excess solution. The resulting solution was divided into two equal parts. One part requires 50 ml of N - for neutralization using phenolphthalein indicator. Another part required 70 ml of N - for neutralization using methyl orange indicator. The value of is
Answer:
2.00
Solution:
Per part, b = mol , c = mol excess . Phenolphthalein: c + b = 0.050; methyl orange: c + 2b = 0.070. So b = 0.02 mol per part, total = 0.04 mol, acid = 0.04/(n+2) mol. Then 0.88 = 0.04/n+2(14n+60) gives 0.88n + 1.76 = 0.56n + 2.4 0.32n = 0.64 = 2. Ans: 2.
Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Double Indicator
⏱ 2mℹ️ Source: QB
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