Question

$2.16\text{g}$ of a metal oxide on strong heating produced $255\text{mL}$ of oxygen at STP (taking 1 atm as standard pressure) and the residue obtained was a pure metal. Calculate the equivalent weight of the metal.

• A. $20\text{g}$
• B. $15\text{g}$
• C. $40\text{g}$
• D. $60\text{g}$

Answer 1

The correct option is C $40\text{g}$

Let the equivalent weight of the metal be $M$
$22400\text{mL}$ volume of $O_2$ has a mass of $32\text{g}$ at STP
$255\text{mL}$ $O_2$ will have a mass of $=\frac{32}{22400}\times 255=0.36\text{g}$

$\text{Mass of the metal}=\text{mass of the oxide}–\text{mass of oxygen}$

$\text{Mass of the metal}=2.16–0.36=1.8\text{g}$

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

$\Rightarrow \frac{1.8}{0.36}=\frac{M}{8}$
On solving,
$M=40\text{g}$