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$40 \mathrm{~mL}$ of $x \mathrm{M} \mathrm{KMnO}_4$ solution is required to react completely with $200 \mathrm{~mL}$ of $0.02 \mathrm{M}$ oxalic acid solution in acidic medium. The value of $x$ is
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$0.04$
$\begin{array}{r}2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5(\mathrm{COOH})_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \\ \mathrm{MnSO}_4+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2\end{array}$
$\begin{aligned} & \therefore \frac{M_1 V_1}{n_1}=\frac{M_2 V_2}{n_2} \\ & \Rightarrow \frac{x \times 40 \mathrm{~mL}}{2}=\frac{0.02 \times 200 \mathrm{~mL}}{5} \\ & \Rightarrow x=0.04 \mathrm{M}\end{aligned}$
$\begin{aligned} & \therefore \frac{M_1 V_1}{n_1}=\frac{M_2 V_2}{n_2} \\ & \Rightarrow \frac{x \times 40 \mathrm{~mL}}{2}=\frac{0.02 \times 200 \mathrm{~mL}}{5} \\ & \Rightarrow x=0.04 \mathrm{M}\end{aligned}$
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