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?
Durga Dibyasingh , 7 Years ago
Grade 11
1 Answers
Arun
Last Activity: 7 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