20 mL of $0.02 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution is used for the titration of 10 mL of $\mathrm{Fe}^{2+}$ solution in the acidic medium. The molarity of $\mathrm{Fe}^{2+}$ solution is $\_\_\_\_$ $\times 10^{-2} \mathrm{M}$. (Nearest Integer)

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Redox Reactions and Electrochemistry

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

____

\times 1 0^{- 2} M

.
(Nearest Integer)

Answer: 24

In acidic medium, the redox reaction is:

Cr X_{2} O X_{7} X^{2 -} + 6 Fe X^{2 +} + 14 H X^{+} 2 Cr X^{3 +} + 6 Fe X^{3 +} + 7 H X_{2} O

**n-factor of

K X_{2} Cr X_{2} O X_{7}

** = 6 (each Cr goes from +6 to +3)
**n-factor of

Fe X^{2 +}

** = 1 (Fe goes from +2 to +3)

Using the equivalence principle:

Equivalents of

K X_{2} Cr X_{2} O X_{7}

= Equivalents of

Fe X^{2 +}

(M \times V \times n)_{K X_{2} Cr X_{2} O X_{7}} = (M \times V \times n)_{Fe X^{2 +}}

0.02 \times 20 \times 6 = M \times 10 \times 1

M = \frac{0.02 \times 20 \times 6}{10} = 0.24 M = 24 \times 1 0^{- 2} M

Therefore, the answer is 24.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer