The volume,in mL,of 0.02 M K_2Cr_2O_7 solut | vedclass

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(D) The balanced redox reaction involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$ and $C_2O_4^{2-}$ to $CO_2$ by $Cr_2O_7^{2-}$ in acidic medium.The $n

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$50$

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(D) The balanced redox reaction involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$ and $C_2O_4^{2-}$ to $CO_2$ by $Cr_2O_7^{2-}$ in acidic medium.The $n$-factor for $K_2Cr_2O_7$ is $6$ $(Cr^{+6} ightarrow Cr^{+3})$.The $n$-factor for ferrous oxalate $(FeC_2O_4)$ is $3$ ($Fe^{+2} ightarrow Fe^{+3}$ and $C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$).Molar mass of $FeC_2O_4 = 56 + 2 	imes 12 + 4 	imes 16 = 144 \ g \ mol^{-1}$.Equating the equivalents: $M_1 	imes V_1 	imes n_1 = \frac{	ext{mass}}{M_w} 	imes n_2$.$0.02 	imes V 	imes 6 = \frac{0.288}{144} 	imes 3$.$0.12 	imes V = 0.002 	imes 3 = 0.006$.$V = \frac{0.006}{0.12} 	imes 1000 \ mL = 50 \ mL$.

The volume,in $mL$,of $0.02 \ M \ K_2Cr_2O_7$ solution required to react with $0.288 \ g$ of ferrous oxalate $(FeC_2O_4)$ in acidic medium is........
(Molar mass of $Fe = 56 \ g \ mol^{-1}$)

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The volume,in $mL$,of $0.02 \ M \ K_2Cr_2O_7$ solution required to react with $0.288 \ g$ of ferrous oxalate $(FeC_2O_4)$ in acidic medium is........(Molar mass of $Fe = 56 \ g \ mol^{-1}$)