Question

On reduction with hydrogen, $3.6\, g$ of an oxide of metal left $3.2\, g$ of metal. If the vapour density of metal is $32$, the simplest formula of the oxide would be

• A. MO
• B. $M_2O_3$
• C. $M_2O$
• D. $M_2O_5$

Answer 1

Answer: (C)

$M_2O$

Answer 2

Vapour density $=\frac{\text { Molecular } \quad \text { Mass }}{2}$ Molecular mass of metal $=2 \times 32=64$ given mass of metal $=3.2 g$ no.of moles of metal $=\frac{3.2}{64}=\frac{1}{20}$ mass of oxygen $=3.6 g -3.2 g =0.4 g$ no.of moles of oxygen $=\frac{3.2}{64}=\frac{1}{40}$ The Metal oxide be in simplest form $M \frac{1}{20} O \frac{1}{40} \Rightarrow M _{2} O$