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

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,

& x=\frac{n}{2m}A \\\ & A=\frac{x\times 2m}{n} \\\ & =\frac{2xm}{n} \\\ \end{aligned}$$ **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.