(20 ~mL ) of \ (Fe^ 2+ ) solution of certain concentration has...

The reaction between \$(\mathrm{Fe}^{2+}$) and \(\mathrm{Cr}}_{2}\mathrm{O}}_{7}^{2-}) (i.e. \(\mathrm{K}}_{2}\mathrm{Cr}}_{2} \mathrm{O}}_{7})) in acidic medium occurs as follows : \$(\mathrm{Cr}}_{2}\mathrm{O}}_{7}^{2-}+14 \mathrm{H}^{+}+6 $\mathrm{Fe}^{2+} \longrightarrow$2 $\mathrm{Cr}^{3+}+$7 \mathrm{H}}_{2}\mathrm{O}+$6$\mathrm{Fe}^{3+}$) The reaction between \$(\mathrm{Fe}^{2+}$) and \(\mathrm{MnO}}_{4}^{-}) (i.e. \$(\mathrm{KMnO}}_{4})) in acidic medium occurs as follows : \$(\mathrm{MnO}}_{4}^{-}+8 \mathrm{H}^{+}+5\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mn}^{2+}+$4$\mathrm{H}}_{2} $\mathrm{O}+$5 $\mathrm{Fe}^{3+}$) (\therefore \mathrm{Cr}}_{2}\mathrm{O}}_{7}^{2-}) react with (6 $\mathrm{Fe}^{2+}$n_1) and \(\mathrm{MnO}}_{4}^{-})react with (5\mathrm{Fe}^{2+}\left(n_2\right)) in a balanced equation. Gram equivalent of \$(\mathrm{Fe}^{2+}) $=$ gram equivalent of \(\mathrm{K}}_{2}\mathrm{Cr}}_{2} \mathrm{O}}_{7})=gram equivalent of \$(\mathrm{KMnO}}_{4}) Therefore, on comparing gram equivalent of \$(\mathrm{K}}_{2}\mathrm{Cr}}_{2} \mathrm{O}}_{7}) and \$(\mathrm{KMnO}}_{4}), we haven_1=6, n_2=5Conc. of \$(\mathrm{K}}_{2}\mathrm{Cr}}_{2} \mathrm{O}}_{7}=0.01\mathrm{M}\left(M_1\right)) Volume of \$(\mathrm{K}}_{2}\mathrm{Cr}}_{2} \mathrm{O}}_{7}=20\mathrm{~mL}\left(V_1\right)) react with \$(\mathrm{Fe}^{2+}) ions (=6 \mathrm{~mol}\left(n_1\right)) Conc. of \(\mathrm{KMnO}}_{4}=) to find (sayM_2) Volume of \$(\mathrm{KMnO}}_{4}=20\mathrm{~mL}\left(V_2\right)) react with \$(\mathrm{Fe}^{2+}) ions (=5 \mathrm{~mol}\left(n_2\right)) Thus, (\quad M_1 \times V_1 \times n_1=M_2 \times V_2 \times n_2) or, (0.01 \times 20 \times 6=M_2 \times 20 \times 5) or \$(M_2=\frac{0.01 \times 6}{5}=$0.012 $\mathrm{M}$) Hence, option (d) is the correct answer.