Question

$1.520g$ of hydroxide of a metal on ignition gave $0.995g$ of oxide. The equivalent weight of metal is

• A. $1.52$
• B. $0.995$
• C. $190$
• D. $9$

Answer 1

Answer: (D)

$\frac{\text{Weight of metal hydroxide}}{\text{Weight of metal oxide}}$
$=\frac{\text{Equivalent weight of metal hydroxide}}{\text{Equivalent weight of metal oxide}}$
or, $\frac{1.520}{0.995}=\frac{E_m+17}{E_m+8}$
[Where $E_m$ =Equivalent weight of metal ]
or, $1.528=\frac{E_m+17}{E_m+8}$
or, $0.528E_m=4.78$
or, $E_m=9$