Question

A metal oxide has the formula ${{X}_{2}}{{O}_{3}}$.It can be reduced by hydrogen to give free metal and water.0.1596g of metal oxide requires 6 mg of hydrogen for complete reduction. The atomic mass of metal in amu is:
(A) 15.58
 (B) 155.8
 (C) 5.58
 (D) 55.8

Answer 1

Hint: The equivalent weight of the metal in the metal oxide ${{X}_{2}}{{O}_{3}}$ can be obtained from the given details in question. As we know equivalent weight is obtained by dividing the atomic weight by valency of the atom. Hence by multiplying the equivalent weight of metal with its valency will give us the atomic mass of metal in amu.

Complete step by step answer:
  -From the given question we could write the chemical reaction as follows
\[{{X}_{2}}{{O}_{3}}+3{{H}_{2}}\to 2X+3{{H}_{2}}O\]

It’s given that six mg of hydrogen reduces 0.1596 g of the given metal oxide ${{X}_{2}}{{O}_{3}}$. Or by converting the mg into gram we can say that $6\times {{10}^{-3}}$ gram of hydrogen reduces 0.1596 gram of metal oxide.
-Now we need to find out the amount of metal oxide reduced by one gram of hydrogen and the calculation is given below
1 g of ${{H}_{2}}$ reacts with $=\dfrac{0.1596}{6\times {{10}^{-3}}}=26.6gm$

Hence the equivalent weight of metal oxide ${{X}_{2}}{{O}_{3}}$ is 26.6 gram. As we know the equivalent weight of metal oxide is the sum of equivalent weights of its constituent particles and we can write it as follows
Equivalent weight of X+ Equivalent weight of O =26.6
Equivalent weight of oxygen = 8
Hence equivalent weight of X =26.6−8
= 18.6
The valency of X in ${{X}_{2}}{{O}_{3}}$ is three. The reason is that it forms an oxide of the form ${{X}_{2}}{{O}_{3}}$ and it belongs to the group three in the periodic table. The group three elements usually have thirteen electrons and in valence shell they have three electrons and as a result the valency of X is 3.
\[Equivalent\text{ }weight=\dfrac{Atomic\text{ }weight}{Valency}\]
\[18.6=\dfrac{Atomic\text{ }weight}{3}\] \[\]
\[Atomic\text{ }weight=18.6\times 3=55.8 amu\]
 Thus the atomic weight of metal X in the metal oxide ${{X}_{2}}{{O}_{3}}$ is 55.8 amu.
So, the correct answer is “Option D”.

Note: The answer can be found through another numerical method also. By looking at the stoichiometry of the reaction we would be able to calculate the number of moles metal oxide reduced and from this we could find the molecular mass of metal oxide. Thus by equating this molecular mass as the sum of masses of its constituent particles we can find the atomic weight of metal.