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The saponification no. of fat or oil is defined as the no. of mg of KOH required to saponify I 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 H2S04 is needed to neutralize excess of KOH. What is saponification no. of peanut oil?

The saponification no. of fat or oil is defined as the no. of mg of KOH required to saponify I 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 H2S04 is needed to neutralize excess of KOH. What is saponification no. of peanut oil?

Grade:11

1 Answers

Arun
25750 Points
6 years ago
Meq of KOH added = 25 * 0.4210 = 10.525
Meq of KOH left = 8.46 *0.2732* 2 = 4.623
Meq of KOH used = 10.525 - 4.623 = 5.902
 
Or 
w * 1000/56 = 5.902
W(KOH) = 0.3305 gm
 
Saponification number = weight of KOH used in mg per gm of oil
= 0.3305 * 1000/1.5763
= 209.6
 
Regards
Arun (askIITians forum expert)

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