Question

$1.70\text{g}$ of a metal oxide required $933\text{mL}$ of hydrogen at STP to be reduced to a pure metal. Find the equivalent weight of the metal.

• A. $12.4\text{g}$
• B. $55.6\text{g}$
• C. $55.6\text{g}$
• D. $67.2\text{g}$

Answer 1

The correct option is A $12.4\text{g}$

Let the equivalent weight of the metal be $M$
$22400\text{mL}$ volume of $H_2$ has a mass of $2\text{g}$ at STP
$933\text{mL}$ will weigh $=\frac{2}{22400}\times 933=0.0833\text{g}$

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

$\Rightarrow \frac{1.70}{0.0833}=\frac{M+8}{1}$

On solving, $M=12.4\text{g}$