A sample of coconut oil weighing 1.5763 g is mixed with 25 mL of 0.4210 M KOH. Some KOH is used in saponification of coconut oil. After the saponification is complete, 8.46 mL of 0.2732 M H_(2)SO_(4) is required to neutralize excess KOH. The saponification number of peanut oil is :

A sample of coconut oil weighing 1.5763 g is mixed with 25 mL of 0.421

1.	A sample of coconut oil weighing 1.5763 g is mixed with 25 mL of 0.4210 M KOH. Some KOH is used in saponification of coconut oil. After the saponification is complete, 8.46 mL of 0.2732 M H_(2)SO_(4) is required to neutralize excess KOH. The saponification number of peanut oil is :
Answer» 209.6 98.9 108.9 218.9 Solution :Number of MILLIEQUIVALENT of KOH added`=25xx0.421=10.525` Number of milliequivalents LEFT unreacted =Number of milliequivalents of `H_(2)SO_(4)` used `= 8.46xx0.2732xx24.623 "" ("Here, basicity of " H_(2)SO_(4)=2)` Number of milliequivalents of KOH used by oil =10.525-4.623=5.902 MASS of KOH used `=(5.902xx56)/(1000)=0.3305 g =330.5 MG` Saponification number = Mass of KOH in mg used by 1 g oil or fat `=(0.3305xx1000)/(1.5763)=209.6`