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

• A

  $\mathrm{M}_2 \mathrm{O}_3$
• Correct answer
  B

  $\mathrm{M}_2 \mathrm{O}$
• C

  $M O$
• D

  $\mathrm{MO}_2$

Solution

Moles of $M=\frac{3.2}{64}=\frac{1}{20} \mathrm{~mol}$ Weight of oxygen $=3.6-3.2=0.4$ Moles of oxygen $=\frac{0.4}{16}=\frac{1}{40}$ $\begin{aligned} & \frac{\text { Moles of metal }}{\text { Moles of oxygen }}=\frac{\frac{1}{20}}{\frac{1}{40}}=2: 1 \\ & \therefore \quad M=2 \text { and } \\ & \therefore \quad \mathrm{O}=1 \text { or the formula of oxide is } M_2 \mathrm{O} . \end{aligned}$ Hence, correct option is (b).

Hence, the correct answer is option: 'B'.