Question

A sample of peanut oil weighing $2$ added to $25$mL of $0.40$M KOH. After saponification is complete, $8.5$mL of $0.28$M $H_{2}S{O}_4$ is needed to neutralize excess of KOH. The saponification number of peanut oil is: (Saponification number is defined as the milligrams of KOH consumed by $1$g of oil).

• A. $146.72$
• B. $223.44$
• C. $98.9$
• D. none of these

Answer 1

Answer: (A)

$146.72$

Reaction of $H_{2}S{O}_4$ with $KOH$:

$H_{2}S{O}_4+2KOH\to K_{2}S{O}_4+2H_{2}O$

Moles of $H_{2}S{O}_4$ used $=\frac{0.28}{1000}\times 8.5=0.00238mol$

For 1 mol of $H_{2}S{O}_4$ , 2 mol of $KOH$ is used, therefore, for 0.00238 mol of $H_{2}S{O}_4$ moles of $KOH$ used$=0.00238\times 2=0.00476mol$

Moles of $KOH$ initially added$=\frac{0.4}{1000}\times 25=0.01mol$

Therefore, moles of $KOH$ consumed $=0.01-0.00476=0.00524mol$

mg of $KOH$ used $=0.00524\times 56\times 1000=293.44mg$

Since saponification number is mg of KOH used to react with 1g of fat, dividing the above weight by 2 we get$=\frac{293.44}{2}=146.72$