Solve this following
Question:

$15 \mathrm{~mL}$ of aqueous solution of $\mathrm{Fe}^{2+}$ in acidic medium completely reacted with $20 \mathrm{~mL}$ of $0.03 \mathrm{M}$ aqueous $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$. The molarity of the $\mathrm{Fe}^{2+}$ solution is Nearest Integer) $\times 10^{-2} \mathrm{M}$ (Round off to the

Solution:

$\mathrm{n}_{\mathrm{eq}} \mathrm{Fe}^{2+}=\mathrm{n}_{\mathrm{eq}} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$

or, $\left(\frac{15 \times \mathrm{M}_{\mathrm{Fe}^{2+}}}{1000}\right) \times 1=\left(\frac{20 \times 0.03}{1000}\right) \times 6$

$\therefore \quad \mathrm{M}_{\mathrm{Ee}^{2+}}=0.24 \mathrm{M}=24 \times 10^{-2} \mathrm{M}$