The saponification number of fat or oil is defined as the number of $m g$ of $K O H$ required to saponify $1 g$ oil or fat. A sample of peanut oil weighing $1.5763 g$ is added to $25 m L$ of $0.421 M K O H$ . After saponification is complete, $8.46 m L$ of $0.2732 M H_{2} S O_{4}$ is needed to neutralize excess of $K O H$ . What is the saponification number of peanut oil?

Open in App

Solution

Molecules equal of $K O H$ added $= 25 \times 0.4210 = 10.525$
Molecules equal of $K O H$ left $= 8.46 \times 0.2732 \times 2 = 4.623$
Molecules equal of $K O H$ used $= 10.525 - 4.623 = 5.902$
Saponification number $=$ Weight of KOH use in mg per gm of oil
$= 0.3305 \times \frac{1000}{1.5763}$
$= 209.6$

Suggest Corrections

Similar questions

2

25

0.40

8.5

0.28

H_{2} S O_{4}

1

The 'saponification value' of an oil or fat is measured in terms of

Which of the following solution is used for saponification of fat or oil?

The saponification of a fat or oil is done using _________ solution for the hot process.

Join BYJU'S Learning Program