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.