20 ~mL of 0.02 \,M \,K_2 Cr_2 O_7 solution is used for the titration of 10 ~mL of Fe^2+ solution in the acidic medium. The molarity of Fe^2+ solution is \underline\hspace2cm \times \,10^-2\, M . (Nearest Integer)

Some Basic Concepts of Chemistry PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q14 with Solution

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Correct answer: 24

In acidic medium, dichromate oxidizes $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ :

\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H}^+ + 6\mathrm{Fe}^{2+} \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{Fe}^{3+} + 7\mathrm{H_2O}

So,

1\text{ mole of }\mathrm{K_2Cr_2O_7} \text{ reacts with } 6\text{ moles of }\mathrm{Fe}^{2+}

Given:

Volume of $\mathrm{K_2Cr_2O_7}$ solution $= 20\,\mathrm{mL} = 0.020\,\mathrm{L}$
Molarity $= 0.02\,\mathrm{M}$

Thus,

\text{moles of }\mathrm{K_2Cr_2O_7} = M \times V = 0.02 \times 0.020 = 4 \times 10^{-4}

Using the stoichiometric ratio $1:6$ ,

\text{moles of }\mathrm{Fe}^{2+} = 6 \times 4 \times 10^{-4} = 24 \times 10^{-4} = 2.4 \times 10^{-3}

Volume of $\mathrm{Fe}^{2+}$ solution $= 10\,\mathrm{mL} = 0.010\,\mathrm{L}$

M_{\mathrm{Fe}^{2+}} = \frac{2.4 \times 10^{-3}}{0.010} = 0.24\,\mathrm{M}

Now express this as:

0.24\,\mathrm{M} = 24 \times 10^{-2}\,\mathrm{M}

The blank is:

\boxed{24}