When 10 , mL of an aqueous solution of KMnO_{ | vedclass

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(C) The balanced redox reaction in acidic medium is: $MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$Equivalents of $KMnO_{4

Vedclass · Accepted Answer

$316$

Vedclass · Answer

(C) The balanced redox reaction in acidic medium is: $MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} ightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$Equivalents of $KMnO_{4} = 	ext{Equivalents of } FeSO_{4}$$n_{factor} 	imes M_{1} 	imes V_{1} = n_{factor} 	imes M_{2} 	imes V_{2}$For $KMnO_{4}$,$n_{factor} = 5$ (reduction of $Mn^{+7}$ to $Mn^{+2}$).For $FeSO_{4}$,$n_{factor} = 1$ (oxidation of $Fe^{+2}$ to $Fe^{+3}$).Given: $V_{1} = V_{2} = 10 \, mL$,$M_{2} = 0.1 \, M$.$5 	imes M_{1} 	imes 10 = 1 	imes 0.1 	imes 10$$M_{1} = \frac{0.1}{5} = 0.02 \, M$Molar mass of $KMnO_{4} = 39 + 55 + (4 	imes 16) = 158 \, g/mol$.Strength in $g/L = Molarity 	imes Molar \, mass = 0.02 	imes 158 = 3.16 \, g/L$.$3.16 \, g/L = 316 	imes 10^{-2} \, g/L$.Thus,the value is $316$.

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