Search any question & find its solution
Question:
Answered & Verified by Expert
On reduction with hydrogen, $3.6 \mathrm{~g}$ of an oxide of metal $(M)$ left $3.2 \mathrm{~g}$ of the metal. If the atomic weight of the metal is 64 . The formula of the oxide is
Options:
Solution:
1731 Upvotes
Verified Answer
The correct answer is:
$\mathrm{M}_2 \mathrm{O}$
Moles of $M=\frac{3.2}{64}=\frac{1}{20} \mathrm{~mol}$
Weight of oxygen $=3.6-3.2=0.4$
Moles of oxygen $=\frac{0.4}{16}=\frac{1}{40}$
$\begin{aligned}
& \frac{\text { Moles of metal }}{\text { Moles of oxygen }}=\frac{\frac{1}{20}}{\frac{1}{40}}=2: 1 \\
& \therefore \quad M=2 \text { and } \\
& \therefore \quad \mathrm{O}=1 \text { or the formula of oxide is } M_2 \mathrm{O} .
\end{aligned}$
Hence, correct option is (b).
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.