Question

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

• A. 1.520
• B. 0.995
• C. 19.00
• D. 9.00

Answer 1

Answer: (D)

$\frac{\text{ Wt. of metal hydroxide }}{\text{ Wt. of metal oxide }}$
$=\frac{\text{ Eq. wt. of metal }+\text{ Eq. wt. of }O{H}^{-}}{\text{Eq. wt. of metal }+\text{ Eq. wt. }{O}^{2-}}$
$=\frac{1.520}{0.995}=\frac{E+17}{E+8}$
$1.520(E+8)=0.995(E+17)$
$1.520E+12.160=0.995E+16.915$
$1.520E-0.995E=16.915-12.16$
or $0.525E=16.915-12.16=4.755$
$E=\frac{4.755}{0.525}=9.0$