Question

The equivalent weight of Cu in CuO and $C{{u}_{2}}O$ (atomic weight of Cu = 63.5) is respectively:
(A) 31.75, 63.5
(B) 63.5, 31.75
(C) 63.5, 63.5
(D) 31.75, 31.75

Answer 1

Hint: Generally, the equivalent weight is used to predict the mass of the substance which reacts with one atom of hydrogen in the acid-base analysis, and also Normality of the solution can be calculated which is defined as the number of the equivalent weight of any substance dissolved in 1L of solution.

Complete step by step solution:
The ratio of the molecular weight of the solute to the valency of the solute is known as equivalent weight. Depending on the type of reaction, the value of the equivalent weight value changed.
The formula of equivalent weight is the molar mass of an atom divided by the number of electrons involved in the reaction per molecule, atom, or ion.
Equivalent weight = $\dfrac{molecular\text{ }weight}{valency}$
Equivalent weight formula for acids,
Equivalent weight of acid = $\dfrac{MW}{B}$
MW = molecular weight of an acid, and B = basicity of acid
Given molecular weight of Cu = 63.5 g
In CuO, the copper element is bivalent which means the valence of Cu is 2.
Therefore, the equivalent weight of Cu in CuO = $\dfrac{molecular\text{ }weight\text{ }of\text{ }Cu}{valency\text{ }of\text{ }Cu}=\dfrac{63.5g}{2}=31.75g$ -- (1)
In $C{{u}_{2}}O$ , the valence of Cu is 1, because its ionic state is $C{{u}^{+}}$ monovalent nature.
Therefore, the equivalent weight of Cu in $C{{u}_{2}}O$ =$\dfrac{molecular\text{ }weight\text{ }of\text{ }Cu}{valency\text{ }of\text{ }Cu}=\dfrac{63.5g}{1}=63.5g$ - (2)
Hence, The equivalent weight of Cu in CuO and $C{{u}_{2}}O$ is respectively 31.75g and 63.5g.

The correct answer is option A.

Note: The use of equivalent weight has largely been superseded by the molar mass in general chemistry. In some reactions, the equivalent weights of the elements can be determined by observing their reaction with oxygen. For the preparation of primary standard solutions in analytical chemistry equivalent weight of compounds plays a key role in the analysis.