Question

A metal oxide has the formula $M_2O_3$. It can be reduced by $H_2$ to give free metal and water. $0.1596\, g$ of $M_2O_3$ 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)

$Z _{2} O _{3}+3 H _{2} \longrightarrow 2 Z +3 H _{2} O$
$\because 6 \times 10^{-3} g H _{2}$ reduces $=0.1596 g$ of $Z_{2} O _{3}$
$\therefore 1 g H _{2}$ reduces $=\frac{0.1596}{6 \times 10^{-3}} gZ _{2} O _{3}$
$=26.6 g$ 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 $=18.6 \times 3=55.8$