Lesson eLearning

A 2.00 g sample of a triglyceride with molecular formula C55H104O6 is completely saponified with excess NaOH. The amount of KOH equivalent required to neutralise the liberated fatty acids is 0.00680 mol. Calculate the saponification value in mg KOH per gram of fat.

medium 3 marks Class 12 · Chemistry · Biomolecules
Answer
Saponification value = 191 mg KOH g⁻¹

The correct answer is Saponification value = 191 mg KOH g⁻¹ — Round to appropriate significant figures: 191 mg KOH g⁻¹.

Step-by-step solution

  1. Write the formula for saponification value: SV = (mol KOH × M(KOH) × 1000) / mass of fat (g).
  2. Substitute the given values: mol KOH = 0.00680 mol, M(KOH) = 56.1 g mol⁻¹, mass = 2.00 g.
  3. Calculate the mass of KOH: 0.00680 mol × 56.1 g mol⁻¹ = 0.38148 g KOH.
  4. Convert this mass to milligrams: 0.38148 g × 1000 = 381.48 mg KOH.
  5. Divide by the mass of the fat sample: 381.48 mg / 2.00 g = 190.74 mg KOH g⁻¹.
  6. Round to appropriate significant figures: 191 mg KOH g⁻¹.
Common mistake: Forgetting to multiply by 1000 to convert grams of KOH to milligrams

Concept tested: Saponification value of a fat

Published · How this solution was prepared and checked

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