Question

A metal oxide has the formula $M_2{O}_3$. It can be reduced by $H_2$ to give free metal and water. $0.1596g$ of $M_2{O}_3$ required 6 mg of $H_2$ for complete reduction. The atomic mass of the metal is

• A. $27.9$
• B. $79.8$
• C. $55.8$
• D. $159.8$

Answer 1

Answer: (C)

$Z_2{O}_3+3H_2⟶2Z+3H_{2}O$
$\because 6\times 10^{-3}g{H}_2$ reduces $=0.1596g$ of $Z_2{O}_3$
$\therefore 1g{H}_2$ reduces $=\frac{0.1596}{6\times 10^{-3}}g{Z}_2{O}_3$
$=26.6g$ of $Z_2{O}_3$
$\therefore$ Equivalent weight of $Z_2{O}_3=26.6$
Equivalent weight of $Z+$ Equivalent weight of $O=26.6$
Equivalent weight of $Z+8=26.6$
Equivalent weight of $Z=(26.6-8)=18.6$
Valency of metal in $Z_2{O}_3=3$
Equivalent weight $=\frac{\text{ Atomic weight }}{\text{ Valency }}$
Atomic weight $=18.6\times 3=55.8$