Question

The oxide of a metal contains 60% of the metal. What will be the approximate percentage of bromine in the bromide of the metal, if the valency of the metal is the same in both the oxide and the bromide?
A.87
B.70
C.77
D.93

Answer 1

Hint: We need to know what metal oxides are and also understand the relation between the valency of a metal and quantity of that metal present in a metal oxide. Metal oxides are solids which are crystalline in nature and consist of a metal cation and an oxide anion. Zinc oxide \[\left( {ZnO} \right)\] is an example of metal oxide. On the other hand, the relation between the valency of a metal and quantity of that metal present in a metal oxide is given by the equivalent weight of that metal which is calculated as atomic weight divided by the valency of the metal.

Complete step by step answer:
As given in the question, a metal oxide contains 60% of the metal. By the definition of metal oxides, it is clear that the metal oxide also contains 40% of oxygen.
Also, equivalent weight is defined as the quantity of a substance which reacts with a fixed quantity of another substance. Therefore, the equivalent weight of the metal will be equal to that of oxygen. The equivalent weight of oxygen is calculated as 8 (atomic weight is 16 and valancy is 8).
Let \[M\] be the equivalent weight of the metal.
Therefore, $\dfrac{{60}}{M} = \dfrac{{40}}{8}$ (since valency of the metal is same in both the oxide and bromine)
$M = 12$
Let B be the percentage of bromine in metal bromide and the equivalent weight of bromide is calculated to be 80.
Therefore, the equivalent weight of bromide in metal bromide will be equal to the equivalent weight of bromine.
i.e. $\dfrac{{100 - B}}{{12}} = \dfrac{B}{{80}}$
or, $B \simeq 87\% $
Therefore, the approximate percentage of bromine in the bromide of the metal is 87
Hence, the correct option is option (A).

Note:
We have to note that the given questions deals with values in percentage amounts hence we see fractions where the percentage of the metal, oxide or bromide is divided by its equivalent weights. For example, $\dfrac{{60}}{M}$ where $M = 12$ ,it means that 12 equivalents of the metal is present in 60% of the metal. The same applies for the oxide and bromide.