A 3.00 g sample of a triglyceride with molecular formula ${\text{C}}_{52}{\text{H}}_{98}{\text{O}}_6$ is completely saponified with excess NaOH; the reaction consumes 0.180 mol of NaOH. Calculate the saponification value in mg KOH per gram of fat.

Step-by-step solution

1. Each mole of NaOH that reacts corresponds to one mole of KOH in the saponification value.
2. Multiply the number of moles of NaOH by the molar mass of KOH to get the mass of KOH released: $0.180\text{mol}\times 56.1{\text{g mol}}^{-1}=10.098\text{g}$.
3. Convert this mass to milligrams: $10.098\text{g}\times 1000=10098\text{mg}$.
4. Divide the milligram amount of KOH by the mass of fat in grams to obtain the saponification value: $10098\text{mg}/3.00\text{g}=3366{\text{mg KOH g}}^{-1}$.

Concept tested: Saponification value calculation

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked