1.

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 H_(2)SO_(4)` is needed to neutralize excess `KOH`. The saponification number of peanut oil is:A. `209.6`B. `108.9`C. `98.9`D. `218.9`

Answer» Correct Answer - A
Meq.of `KOH` added `=25xx0.4210=10.525`
Meq. of `KOH` left `=8.46xx0.2732xx2=4.623`
`:.` Meq. of `KOH` used by oil`=10.525-4.623=5.902`
or `(w)/(56)xx1000=5.902`
or `w_(KOH)=0.3305g`
`:.` Saponification no. `=wt.of KOH` used in `mg per g` of oil
`=(0.3305)/(1.5763)xx1000=209.6`


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