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A sample of peanut oil weighing 1.5763 g...

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 H_(2)SO_(4)` is needed to neutralize excess `KOH`. The saponification number of peanut oil is:

A

146.72

B

223.44

C

98.44

D

98.9

Text Solution

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The correct Answer is:
A
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