Experimentally, it was found that a metal oxide has formula M0.98O1. Metal M is present as M2+ and M3+ in its oxide. The percentage of the metal which exists as M3+ would be.

Let number of M3+ ions is ‘x’, then the number of M2+ ions will be (0.98 - x). Net charge of oxide will be zero. ∴+2(0.98-x)+3(x)+(-2)1=01.96+x-2=0x=0.04Percentage of M3+=0.040.98×100=4.08