\(n=\frac{\text { Mass given }}{\text { Molar mass }}\)
\(0.01186=\frac{1}{M} \Rightarrow M=84.3\, \mathrm{g\,mol}^{-1}\)
$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
અહી $20\, mL$ of $0.1\, M\, KMnO_4$ એ કોના બરાબર હશે