Question

$80mL$ of $KMn{O}_4$ solution reacts with $3.4g$ of $N{a}_2{C}_2{O}_4.2H_{2}O$ in an acidic medium. The molarity (in M) of the $KMn{O}_4$ solution is:

Answer 1

Redox reactions involved are as follows,
$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$ -(eq.1)
$C_2{O}_4^{2-}⟶2C{O}_2+2e^{-}$ -(eq.2)
now, multiplying (eq.1) by 2 and (eq.2) by 5 and adding the two we get:
$2Mn{O}_4^{-}+16H^{+}+5C_2{O}_4^{2-}⟶2M{n}^{2+}+10C{O}_2+8H_{2}O$

Molar mass of $N{a}_2{C}_2{O}_4.2H_{2}O=170g/mol$
So moles of $N{a}_2{C}_2{O}_4.2H_{2}O=\frac{3.4}{170}=0.02$
And from the above balanced equation we can see that 5 moles of $C_2{O}_4^{2-}$ reacts with 2 moles of $Mn{O}_4^{-}$, so 0.02 moles of $C_2{O}_4^{2-}$ will react with $=\frac{2}{5}\times 0.02=0.008$ moles of $Mn{O}_4^{-}$
So molarity of $KMn{O}_4$ solution (M) is,
$M=\frac{moles}{\text{volume of solution}}$
$M=\frac{0.008}{80\times {10}^{-3}}$
$M=0.1$