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JEE Main Chemistry question

ChemistryRedox ReactionsJEE Main

A sample of peanut oil weighing 1.5763 g is added to 25 mL of 0.4210 M KOH. After saponification is complete, 8.46 mL of 0.2732 M HSO is needed to neutralise excess KOH. The saponification number of peanut oil is:

Answer: (A)

Solution

mmol KOH initial . mmol HSO, neutralising excess KOH mmol. KOH used

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Question ID VK-130936