The saponification number of fat or oil is defined as the number of mg of KOH required to saponify 1 g oil or fat. A sample of mustard oil weighing 2.0 g is added to 25.0 ml of 0.5 M - KOH . After sap

Question

The saponification number of fat or oil is defined as the number of mg of $KOH$ required to saponify 1 g oil or fat. A sample of mustard oil weighing 2.0 g is added to 25.0 ml of 0.5 M - $KOH$ . After saponification is complete, 10.0 ml of 0.25 M - $H_{2} S O_{4}$ is needed to neutralize excess of $KOH$ . What is the saponification number of the oil?

Answer:

210.00

Solution:

$KOH$ taken = $25 \times 0.5$ = 12.5 mmol. Excess = 2×(10×0.25) = 5 mmol. Consumed = 7.5 mmol = $7.5 \times 56$ = 420 mg per 2.0 g, i.e. 210 mg per g. Ans 0210.

TestHub · Accepted Answer

KOH taken = 25 × 0.5 = 12.5 mmol. Excess = 2×(10×0.25) = 5 mmol. Consumed = 7.5 mmol = 7.5 × 56 = 420 mg per 2.0 g, i.e. 210 mg per g. Ans 0210.

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