Question

The equivalent mass of metal M is x and the formula of its oxide is ${{M}_{n}}{{O}_{m}}$ . The atomic mass of metal is:
A. $\frac{2n-x}{m}$
B. $\frac{2mx}{n}$
C. \[\frac{2mn}{x}\]
D. \[\frac{mn}{x}\]

Answer 1

Hint: Atomic mass is basically equal to the number of protons and neutrons in an atom.
Atomic mass is related to the equivalent mass by the formula:$Equivalent\text{ }mass=\frac{Atomic\text{ }mass}{Valency}$

Complete answer:
- We are being provided with an equivalent mass of metal M as x and the formula of oxide is ${{M}_{n}}{{O}_{m}}$. Let’s discuss about the solution:
- We need n moles of M and m/2 moles of ${{O}_{2}}$to form ${{M}_{n}}{{O}_{m}}$.
\[nM+\frac{m}{2}{{O}_{2}}\to {{M}_{n}}{{O}_{m}}\]
- Let, the atomic mass of metal=${{A}_{g}}$,
 And atomic mass of oxygen is 32,
- So, we get:
\[n\times {{A}_{g}}\text{ }of\text{ }M+\frac{m}{2}\times 32g\text{ }of\text{ }{{O}_{2}}\to {{M}_{n}}{{O}_{m}}\]
8gm of oxygen combined with metal will be its equivalent weight x.
\[\frac{m}{2}\times 32g\text{ }of\text{ }{{O}_{2}}\to n\times {{A}_{g}}\text{ }of\text{ }M\]
As we can see that equivalent weight (x) of oxygen is 8g that will combine with metal:
\[x=8g\text{ }of\text{ }{{O}_{2}}\to \frac{n}{2m}\times {{A}_{g}}\text{ }of\text{ }M\]
So, by solving we get,
\[\begin{align}
  & x=\frac{n}{2m}A \\
 & A=\frac{x\times 2m}{n} \\
 & =\frac{2xm}{n} \\
\end{align}\]

Hence, we can conclude that the correct option is (B), that is the atomic mass of metal is: $\frac{2mx}{n}$

Additional information:
- Metal oxides are crystalline solids which contain an oxide anion and a metal cation. These react with water and form bases. Thus, these metal oxides are also called basic oxides.

Note:
- We should not get confused in terms of equivalent mass and atomic mass.
- Atomic mass is the sum of the number of neutrons and protons present in the nucleus of an atom. Whereas, equivalent mass is the mass of a given substance which combines with a fixed quantity of another substance.