$p$ avoid bodical

Q$3$: $10mL$ of $0\mathrm{.}05$MKOH was required to neutralize free fatty acid of $0\mathrm{.}202g$ of an oil sample $($equivalent weight of oleic acid $=282\frac{g}{m}ol.).$ Calculate the acid value of this sample. Titration of blank takes $1\mathrm{.}1mL$ of the same solution of KOH $(4$ marks$).$

$\frac{V_1-\frac{V_{2}N2821\text{d}\text{o}\text{t}\text{s}}{\text{d}\text{o}\text{t}\text{s}1\text{d}\text{o}\text{t}\text{s}}}{1\text{d}\text{o}\text{t}\text{s}}$