15 $m L$ of an aqueous solution of $F e^{2 +}$ in the 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 _______ $\times 10^{- 2}$ $M$ . (Round off to the Nearest Integer).

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Solution

Given solution of $F e^{2 +}$ in the acidic medium completely reacts with given aqueous solution of $C r_{2} {O_{7}}^{2 -}$ .
Hence, by law of equivalence,
$M_{e q}$ of $F e^{2 +}$ = $M_{e q}$ of $C r_{2} {O_{7}}^{2 -}$
$M \times 15 \times n - f a c t o r = 0.03 \times 20 \times n - f a c t o r$
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 electron is involved. Hence, its n-factor is 6.
$∴$
$M o l a r i t y = 0.24 M \Rightarrow 24 \times 10^{- 2} M$

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