. 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
Related questions
