9.5 g FeSO₄ কে জারিত করতে 1 M KMnO₄ দ্রবণের কত mL প্রয়োজন?

for $\mathrm{FeSO}_{4}$,

$\begin{array}{l} e_{1}=1 \quad M=152 \\ W_{1}=9-5 \end{array}$

for $\mathrm{KMnO}_{4}$,

$e_{2}=5 \quad S_{2}=1$

$\begin{aligned} & e_{1} V_{1} S_{1}=e_{2} V_{2} S_{2} \\ \Rightarrow & 1 \times V_{1} \times \frac{W_{1} \times 1000}{M \times V_{1}}=5 \times V_{2} \times 1 \\ \Rightarrow & \frac{9.5 \times 1000}{152}=5 \times V_{2} \times 1 \\ \therefore & V_{2}=12.5 \mathrm{ml}\end{aligned}$