$15 m L$ of aqueous solution of ${F e}^{2 +}$ in acidic medium completely reacted with $20 m L$ of $0.03 M$ aqueous ${C r}_{2} O_{7}^{2 -}$ . The molarity of the ${F e}^{2 +}$ solution is $10^{- 2} M$ . (Round off to the Nearest Integer).

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answer is 24.

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Detailed Solution

$M_{e q}$ of ${F e}^{2 +} = M_{e q}$ of ${C r}_{2} O_{7}^{2 -}$

$M \times 15 \times n -$ factor $= 0.03 \times 20 \times n -$ factor

In the reaction, ${F e}^{2 +}$ is oxidised to ${F e}^{3 +}$ hence, the $n$ -factor is 1.

Also, ${C r}_{2} O_{7}^{2 -}$ is reduced to $2 {C r}^{3 +}$ . Thus, 6 electrons are involved. Hence, its $n$ -factor is 6.

Molarity $= 0.24 M \Rightarrow 24 \times 10^{- 2} M$

Hence the correct answer is 24.

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