calculate the volume of  0.05 KMno4 solution required to oxidised completely 2.70 g of oxalic acid in acidic solution​

Dear Student,


To calculate number of moles of KMnO4 required to oxidised completely 2.70 g of oxalic acid in acidic medium.

Molecular weight of H₂C₂O₄.2H₂0 = 126 g
$\text{Number of moles of oxalic acid = }\frac{\mathrm{Mass} \mathrm{of} \mathrm{oxalic} \mathrm{acid}}{\mathrm{Molar} \mathrm{mass}}\text{ =}\frac{2.70 }{126 }\text{ =0.0214 g}$
$$\begin{gathered} {\text{5 moles of oxalic acid is oxidised by 2 moles of KMnO}}_{\text{4}} \\ \text{So, 0.021 moles }\mathrm{of} \mathrm{oxalic} \mathrm{acid} \mathrm{is} \mathrm{oxidised} \mathrm{by}\frac{ 2}{5}\times 0.021 \mathrm{moles} \mathrm{of} {\mathrm{KMnO}}_{\text{4}} =0.00857 \end{gathered}$$
To calculate volume of 0.05 M  KMnO₄ solution :
$$\begin{gathered} {\text{0.05 moles of KMnO}}_{\text{4 }}\text{ are contained in 1000 ml of the solution} \\ \mathrm{So}. 0.00857 \mathrm{moles} \mathrm{of} {\mathrm{KMnO}}_{\text{4 }} \mathrm{are} \mathrm{contained} =\frac{ 1000 }{0.05}\times 0.00857 \mathrm{ml} =171.42 \mathrm{ml} \end{gathered}$$