Question

In an experiment, \[4g\] of ${M_2}{O_x}$ oxide was reduced to $2.8g$ of the metal. If the atomic mass of the metal is $56gmo{l^{ - 1}}$. The number of $O - $atoms in the oxide is:
A. 1
B. $\dfrac{{48}}{{16}}$
C. $\dfrac{{58}}{{32}}$
D. 4

Answer 1

Hint: We can calculate the number of $O - $atoms in the oxide by using the mass of metal oxide, reduced mass of metal, atomic mass of the metal and mass of oxygen.

Complete step by step answer:
Given data contains,
Mass of ${M_2}{O_x}$ oxide is $4g$.
Reduced mass of the metal is $2.8g$.
Atomic mass of the metal is $56g/mol$.
We know that $16g/mol$ is the atomic mass of the oxygen.
The metal oxide ${M_2}{O_x}$ was reduced to metal $M$ and oxygen. We can write this equation as,
${M_2}{O_x} \to 2M + x{O_2}$
The number of oxygen atoms is nothing but the valency of the metal.
$4g$ of the metallic oxide is reduced to $2.8g$ of the metal.
$4g \to 2.8g$
Let us keep the number of oxygen atoms as x.
Equivalent weight of the metal oxide ${M_2}{O_x}$ would be equal to equivalent weight of the metal M.
${M_2}{O_x} \to 2M$
$2 \times 56 + x \times 16 \to 2 \times 56$
$4 \times 2 \times 56 = 2.8\left( {2 \times 56 + x \times 16} \right)$
On solving the value of x, we get,
$x = \dfrac{{48}}{{16}}$

On simplifying, we get the value of x as 3. So, the number of oxygen atoms is three and the valency of the metal is also 3.
So, the formula of metal oxide would be ${M_2}{O_3}$.

So, the correct answer is Option b.

Note:
An alternate approach to solve this problem is,
We know that metal oxide is reduced to M. The equation is,
${M_2}{O_x}\xrightarrow{{{\text{Reduction}}}}M$
The equivalent weight of metal oxide ${M_2}{O_x}$ is equal to the equivalent weight of the metal.
Given data contains,
Mass of ${M_2}{O_x}$ oxide is $4g$.
Reduced mass of the metal is $2.8g$.
Atomic mass of the metal is $56g/mol$.
We know that $16g/mol$ is the atomic mass of the oxygen.
Let us keep the number of oxygen atoms as x. We can calculate the number of oxygen atoms as,
$\dfrac{{{\text{Weight of }}{{\text{M}}_2}{{\text{O}}_x}}}{{{\text{Equivalent weight of }}{{\text{M}}_2}{{\text{O}}_x}}} = \dfrac{{{\text{Weight of metal}}}}{{{\text{Equivalent weight of metal}}}}$
Substituting the known values we get,
\[\dfrac{4}{{\dfrac{{2 \times 56 + x \times 16}}{{2x}}}} = \dfrac{{2.8}}{{\dfrac{{56}}{x}}}\]
We can solve the above expression as,
$\dfrac{4}{{56 + 8x}} = \dfrac{{2.8}}{{56}}$
We can simplify the expression as,
$ \Rightarrow \dfrac{1}{{14 + 2x}} = \dfrac{1}{{20}}$
$ \Rightarrow 2x = 6$
$ \Rightarrow x = 3$
Therefore, the number of oxygen atoms is three, and the formula of the oxide is ${M_2}{O_3}$.