The saponification no. of fat or oil is defined as the no. of mg of KOH required to saponify 1 g oil or
fat. 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 H2SO. is needed to neutralize excess of KOH. What
is saponification no. of peanut oil?
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Molecules equal of KOH added =25×0.4210=10.525
Molecules equal of KOH left=8.46×0.2732×2=4.623
Molecules equal of KOH used =10.525−4.623=5.902
Saponification number = Weight of KOH use in mg per gm of oil
=0.3305× 1.5763 /1000
=209.6
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