Question

A sample of peanut oil weighing 1.5763g is added to 25mL of 0.4210M KOH. After saponification is complete 8.46mL of 0.2732M H₂SO₄ is needed to neutralize excess KOH. The saponification number of peanut oil is:

A. 209.6
B. 108.9
C. 98.9
D. 218.9

Answer 1

The saponification number of peanut oil can be calculated using the given data. The saponification number is 209.6. Therefore, the correct option is A.

Step-by-step explanation:

The saponification number of peanut oil can be calculated using the given data. The saponification number is the number of milligrams of KOH required to saponify the free and combined fat in 1 gram of the oil. We can use the equation:

Mass of KOH = (Volume of H2SO4 × Molarity of H2SO4) / Molarity of KOH

By substituting the given values, the saponification number of peanut oil is 209.6.