Chemistry - Redox Reactions Question with Solution | TestHub
ChemistryRedox ReactionsProblem involving Equivalent ConceptHard2 minQB
ChemistryHardnumerical
The saponification number of fat or oil is defined as the number of mg of KOH required to saponify 1 g oil or fat. A sample of peanut oil weighing 1.5 g is added to 25.0 ml of 0.4 M - KOH. After saponification is complete, 8.0 ml of 0.25 M - H₂SO₄ is needed to neutralize excess of KOH. What is the saponification number of peanut oil?
Answer:
224.00
Solution:
KOH taken = 25×0.4 = 10 mmol. Excess KOH = 2×(8.0×0.25) = 4 mmol. KOH consumed = 6 mmol = 6×56 = 336 mg per 1.5 g oil. Per gram: 336/1.5 = 224 mg. Saponification number = 0224.
Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Problem involving Equivalent Concept
⏱ 2mℹ️ Source: QB
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