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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` |
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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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