Experimentally it was found that a metal oxide has formula M 0.98 O. Metal M, is present as M 2+ and M 3+ in its oxide. Fraction of the metal which exists as M 3+ would be: (1) 4.08% (2) 6.05% (3) 5.08% (4) 7.01%

M+2 and M+3 combinely form M0.98 So, let no. Of M+3 be x then M+2 should be 0.98-xx+[(0.98-x)×2]= 2 (since valency of O is 2)So after this simple calculation, we get M+3=0.04 % of M+3= 0.04/0.98×100= 4.08%That`s all guys!!!!

Dear Student

Given
Metal oxide = M_0.98O
If ‘x’ ions of M are in +3 state, then
3x + (0.98 – x) × 2 = 2
x = 0.04
So the percentage of metal in +3 state would be 0.04/.98* 100 = 4.08%

I hope this answer will help you.