Search any question & find its solution
Question:
Answered & Verified by Expert
$1.520 \mathrm{~g}$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is:
Options:
Solution:
2429 Upvotes
Verified Answer
The correct answer is:
9
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]
$$
\begin{array}{l}
\Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\
\Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\
\therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9
\end{array}
$$
$$
\text { So, } \frac{E+17}{E+8}=\frac{1.52}{0.995}
$$
[17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen]
$$
\begin{array}{l}
\Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\
\Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\
\therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.