$100 mL$ of $0.1 M {KMnO}_{4}$ is consumed in its titration with oxalic acid in presence of dil. $HCl$ whose excess amount was taken. The ${Cl}_{2} ()$ produced is reacted with excess of KI solution producing $I_{2}$ which required $170 mL$ of $0.2$ hypo solution for complete reaction. Calculate millimoles of oxalic acid consumed.

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answer is 8.

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Detailed Solution

In the following reaction:

$\begin{array}{l} {Cl}_{2} + KI ⟶ I_{2} + KCl \\ 2 \times {mmCl}_{2} = 2 mmI \\ I_{2} + {Na}_{2} S_{2} O_{3} ⟶ {Na}_{2} S_{4} O_{6} \\ 2 \times {mmI}_{2} = 1 \times mm of Hypo \end{array}$

$\begin{array}{l} 2 \times {mmI}_{2} = 1 \times 170 \times 0.2 \\ {mmI}_{2} = 17 \\ mmof {Cl}_{2} = 17 \\ {KMnO}_{4} + H_{2} C_{2} O_{4} ⟶ \\ 5 \times (10 - \frac{34}{5}) = 2 \times {mmH}_{2} C_{2} O_{4} \\ 5 \times \frac{16}{5} \times \frac{1}{2} = 8 {mmH}_{2} C_{2} O_{4} \\ {KMnO}_{4} + HCl ⟶ {Cl}_{2} \\ 5 \times mm {KMnO}_{4} = 2 \times 17 \\ KmKMnO = \frac{34}{5} \end{array}$