28 ml of 0.1 m oxalic acid solution requires 10 ml of KMnO4 for titration.10 ml of this sample of KMnO4 when added to an excess of NH2OH liberates N2 at stp. volume of N2 liberated at STP?

Given

28 ml of 0.1 m oxalic acid solution requires 10 ml of KMnO4 for titration. 10 ml of this sample of KMnO₄ when added to excess of NH₂OHliberates N2 at STP.

Find out

We need to find out the volume of N₂ liberated at STP.

Solution

In acidic medium MnO4^– →Mn⁺²

So (NV)_KMnO4 = (NV)_H2C2O4
N × 10 = 28 × 0.1 × 2
N = 0.56

Molarity KMnO4 in acidic medium = 0.56 / 5 with NH₂OH (medium is weakly basic).

(NV)_KMnO4⁼ W / E X 1000

( 0.56 / 5 X 3 ) X 10= W / M/2 X 1000

mole of N₂ = 1.68 × 10^–3

Volume at STP = 1.68 × 10^–3 × 22400 = 37.63 = 38 ml.

Answer

Volume of N₂ at STP= 38 ml.