A 10.0 g sample of triglyceride $C_{57} H_{104} O_{6}$ is completely saponified with excess NaOH; the reaction consumes 0.375 mol of NaOH. Calculate the saponification value (mg KOH g⁻¹).

hard 4 marks Class 12 · Chemistry · Biomolecules

Answer

2104 mg KOH g⁻¹

The correct answer is 2104 mg KOH g⁻¹ — Hence the saponification value is 2104 mg KOH g⁻¹.

Step-by-step solution

Determine the molar mass of the triglyceride: $M = 57 (12.01) + 104 (1.008) + 6 (16.00) = 886.4$  g mol⁻¹.
Each mole of NaOH neutralises one mole of fatty acid; therefore 0.375 mol NaOH corresponds to 0.375 mol KOH for the value calculation.
Calculate grams of KOH: $0.375 mol \times 56.11 g mol^{- 1} = 21.04$  g.
Convert to milligrams: $21.04 g = 21040$  mg.
Divide by fat mass in grams to get saponification value: $21040 mg /10.0 g = 2104$  mg KOH g⁻¹.
Hence the saponification value is 2104 mg KOH g⁻¹.

Common mistake: Using NaOH moles directly as mg KOH or incorrect molar mass

Concept tested: Saponification value calculation