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 :

Options

A. 209.6
B. 98.9
C. 108.9
D. 218.9

Answer

A

Explanation

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`