A 5.00 g sample of a pure triglyceride (molecular formula C₅₅H₁₀₄O₆) is completely saponified with excess NaOH. The reaction consumes 0.250 mol of NaOH. Calculate the saponification value of the fat in mg KOH g⁻¹.
Answer
2.81 × 10³ mg KOH g⁻¹
The correct answer is 2.81 × 10³ mg KOH g⁻¹ — Round the result to three significant figures: 2.81 × 10³ mg KOH g⁻¹.
Step-by-step solution
- Write the relation between NaOH and KOH for saponification: 1 mol NaOH corresponds to 1 mol KOH in terms of neutralising capacity.
- Calculate the mass of KOH equivalent to 0.250 mol NaOH: mass KOH = 0.250 mol × (56.11 g mol⁻¹) = 14.0275 g.
- Convert this mass to milligrams: 14.0275 g = 14027.5 mg.
- Determine the amount of KOH required per gram of fat: saponification value = 14027.5 mg ÷ 5.00 g = 2805.5 mg g⁻¹.
- Round the result to three significant figures: 2.81 × 10³ mg KOH g⁻¹.
Common mistake: Using the mass of NaOH directly instead of converting to the equivalent mass of KOH.
Concept tested: Saponification value calculation
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