Question

On heating $0.199\text{g}$ of a metallic oxide in a stream of hydrogen, $0.045\text{g}$ of water is formed. Find the equivalent weight of the metal.

• A. $61.2\text{g}$
• B. $31.8\text{g}$
• C. $12.7\text{g}$
• D. $15.5\text{g}$

Answer 1

The correct option is B $31.8\text{g}$

Given: weight of the metallic oxide $=0.199\text{g}$
$16\text{g}$ of oxygen is present in $18\text{g}$ of $H_{2}O$
So, $O$ present in $0.045\text{g}{H}_{2}O=\frac{16}{18}\times 0.045=0.04\text{g}$
Weight of the metal $=0.199–0.04=0.159\text{g}$

$\frac{\text{weight of the metal oxide}}{\text{weight of the metal}}=\frac{\text{equivalent weight of the metal oxide}}{\text{equivalent weight of the metal}}$

$\Rightarrow \frac{0.199}{0.159}=\frac{M+8}{M}$
On solving, $M=31.8\text{g}$