$1.520 \mathrm{~g}$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent…
$1.520 \mathrm{~g}$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is:
1.52
0.995
190
9
Solution
Let $E$ be the equivalent weight of the metal
$
\text { So, } \frac{E+17}{E+8}=\frac{1.52}{0.995}
$
[17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen]
$
\begin{array}{l}
\Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\
\Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\
\therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9
\end{array}
$