On reduction with hydrogen $3.6 \mathrm{~g}$ of an oxide of metal leaves $3.2 \mathrm{~g}$ of metallic…

On reduction with hydrogen $3.6 \mathrm{~g}$ of an oxide of metal leaves $3.2 \mathrm{~g}$ of metallic residue. If the atomic mass of metal is 64 , the formula of metal oxide is
  1. $\mathrm{M}_{2} \mathrm{O}_{3}$
  2. $\mathrm{M}_{2} \mathrm{O}$
  3. $\mathrm{MO}$
  4. $\mathrm{MO}_{2}$

Solution

Mass of oxygen which gets displaced from metal oxide $=0.4 \mathrm{~g}$
Now, $0.4 \mathrm{~g}$ of oxygen combines with metal $=3.2 \mathrm{~g}$
$8 \mathrm{~g}$ of oxygen combines with metal $\frac{3.2}{0.4} \times 8=64 \mathrm{~g}$
GEW of metal $=64 \mathrm{~g}$
Valency of metal $=\frac{\mathrm{GAM}}{\mathrm{GEW}}=\frac{64 \mathrm{~g}}{64 \mathrm{~g}}=1$
Hence, formula of oxide is $\mathrm{M}_{2} \mathrm{O}$.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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