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. $M_{2}O_{3}$
• C. $M_{2}O$
• D. $M_{2}O_{5}$

Answer 1

Answer: (C)

$M_{2}O$

Answer 2

First find equivalent weight, then molecular weight and then molecular formula by using various formulas. Equivalent weight of metal $=\frac{\text { weight of metal }}{\text { weight of oxygen }} \times 8$ Given, weight of oxide of metal $=3.6 g$ Weight of metal $=3.2 g$ Vapour density of metal $=32$ $\therefore$ Equivalent weight of metal $=\frac{3.2}{3.6-3.2} \times 8$ $=\frac{3.2}{0.4} \times 8=64$ Molecular weight $=2 \times$ vapour density $=2 \times 32=64 n$ $=\frac{\text { molecular weight }}{\text { equivalent weight }}$ $=\frac{64}{64}=1$ Let the formula of metal oxideb$=M_{2} O_{n}$ $\therefore$ Formula of metal oxide $=M_{2} O$ $(\because n=1)$