. A sample of peanut oil weighing


1.576 g is added to 25 ml of 0.421 M KOH. After saponification is complete, 8.46 ml of 0.273 M H2SO4 is needed to neutralize excess of KOH. The saponification number of peanut oil is (saponification number is defined as the milligrams of KOH consumed by 1 g of oil)

A. 209.6 

B. 108.9

C. 98.9 

D. 218.9


 

 

Best Answer

Ans: A

Sol: Equivalents of KOH used by oil = [  25 x 0.421 -8.46 x 0.73 x 2[ x 10-3

Moles of KOH used = 5.90 ´ 10–3

Mass of KOH used in milligrams = 5.90 ´ 10–3 ´56 ´ 1000 = 330.40 

Saponification number = 330.40/1.576= 209.64

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