On reduction with hydrogen, 3.6 ~g of an oxide of metal (M) left 3.2 ~g of the metal. If the atomic weight of the metal is 64 . The formula of the oxide is

Step-by-step solution

Moles of M= 3.2 64 = 1 20 ~mol Weight of oxygen =3.6-3.2=0.4 Moles of oxygen = 0.4 16 = 1 40 aligned & Moles of metal Moles of oxygen = 1 20 1 40 =2: 1 & M=2 and & O =1 or the formula of oxide is M₂ O . aligned Hence, correct option is (b).