Given:

Oil sample $=1.5g$

Normality of $\text{KOH}=0.5N$

Normality of $\text{HCl}=0.5N$

Blank tiration $=25ml$

Formula:

$\text{Saponification value}(mg/g)=[\text{(Blank level)-(Titration Volume)}]\times \text{Normality of KOH}\times \text{Equivalent weight of KOH}$

Solution:

$\begin{align*} \text{Saponification value}(mg/g) &=\dfrac{[(25ml)-(15ml)]\times 0.5N\times 56}{1.5}\\[2ex] &=\dfrac{10\times 0.5\times 56}{1.5}\\[2ex] &=\dfrac{280}{1.5}\\[2ex] &=186.67\end{align*}$

Therefore, Saponification value is 186.67