Chemistry - Redox Reactions Question with Solution | TestHub
ChemistryRedox ReactionsTitrationMedium2 minQB
ChemistryMediumnumerical
A mixture containing and requires 20 ml of 0.05 N iodine for titration. The resulting solution is then acidified and excess of was added. The liberated iodine required 0.992 g hypo, , for complete reaction. The reactions are + + + + ; + + + + . The mole per cent of in the original mixture is (As = 75)
Answer:
25.00
Solution:
First titration: = 1 meq = 0.5 mmol, so = 0.25 mmol. Hypo = 0.992/248 = 4 mmol, so liberated = 2 mmol and total = 1.0 mmol. Original = 1.0-0.25 = 0.75 mmol. Mole % = 0.25/1.0×100 = 25%. Ans 0025.
Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Titration
⏱ 2mℹ️ Source: QB
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