Question

5 gm of oil was saponified with 50 ml of 0.5 alcoholic KOH. After refluxing

for 2 hours , the mixture was titrated by 15 ml of 0.5 N HCl Find

saponification value
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Answer 1

Answer:

Sap value = (Blank titration reading – Back titration reading) X N KOH X 56 /weight of oil

= (50 -15) x 0.5 x 56 / 5

= 196 mg of KOH /gm of oil

Explanation:

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