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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. 146.72B. 223.44C. 98.44D. 98.9 |
Answer» Correct Answer - A |
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