Let, the number of M2+ ions =x Then, the number of M3+ ions will be 0.96−x We know, the overall charge in the metal oxide is zero. So, x(2)+(0.96−x)(3)+1(−2)=0 ⇒2x+2.88−3x=2 ⇒−x=−0.88 ⇒x=0.88 ∴ Number of M3+ ions =0.96−0.88=0.08 ∴ Percentage of M3+ ions =