- Download PDF
1 Answer | Add Yours
40 ml of `KMnO_4` is required to oxidize 0.280 g of `Na_2C_2O_4` . When the two solutions are mixed the reaction that takes place is `C_2O_4 + MnO_4 -> 2CO_2 + MnO_2 + O_2` . The `K^+` and `Na^+` ions react with the cations from the acid to form salts.
From the equation provided one mole of `C_2O_4` and one mole of `MnO_4` react with each other.
The molar mass of `Na_2C_2O_4` is 134, 0.280 g of `Na_2C_2O_4` is equivalent to `7/3350` moles. As this reacts with `MnO_4` present in 40 ml of `KMnO_4` , the molarity of the `KMnO_4` solution is `7/134 ~~ 0.0522`
The correct answer is option D.
We’ve answered 327,786 questions. We can answer yours, too.Ask a question