Chapter 23, Problem 23.86CP

Interpretation Introduction

Interpretation:

The number of grams of $\text{NaOH}$ is needed to saponify $\text{5}\text{.0}\text{g}$ of the oil has to be calculated.

Concept Introduction:

Saponification index:

The number of grams of $\text{NaOH}$ is needed to saponify one gram of the particular fat is known as Saponification index.

${\text{m}}_{{\text{NaOH}}_{\text{min}}}\text{=}\frac{\text{m}}{\text{M}}\text{ × }\frac{\text{n}\text{mol}\text{NaOH}}{\text{1}\text{mole}\text{triglyceride}}{\text{ × M}}_{\text{NaOH}}$

Where,

m is the number of grams of oil

M is the average molar mass of the sample

n depends on the number of fatty acids $\text{(3)}$ attached to the glycerol

${\text{M}}_{\text{NaOH}}$ is the molar mass of the sodium hydroxide.