Question

In the reaction : $Mn{{O}_{2}}+4HCl\to MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O$, the equivalent weight of HCl is:
(A) M
(B) $\dfrac{M}{2}$
(C) 2M
(D) $\dfrac{M}{4}$

Answer 1

Hint: Equivalent weight has the dimensions and units of mass, unlike atomic weight, which is dimensionless. The equivalent weight of a compound can be calculated by dividing the molecular mass by the number of positive or negative electrical charges that result from the dissolution of the compound.

Complete step by step answer:
Equivalent weight (also known as gram equivalent) is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance. The equivalent weight of an element is the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. These values correspond to the atomic weight divided by the usual valence.
The n factor of 4 moles of HCl is 2, since there is change of two electron from HCl to $C{{l}_{2}}$, therefore, for one moles of HCl, the n-factor will be equal to $\dfrac{1}{2}$.
As per question, if M is the molecular weight of HCl then,Equivalent weight of HCl will be,
\[\begin{align}
 & \text{Equivalent weight=}\dfrac{\text{molecular weight}}{n-factor} \\
 & \text{Equivalent weight=}\dfrac{M}{\dfrac{1}{2}} \\
 & \text{Equivalent weight=}2M \\
\end{align}\]
So, the correct answer is “Option C”.

Note: n-factor is atomic weight or molecular weight by equivalent weight. The n factor for acids is its basicity, that is, the number of replaceable hydrogen ions per molecule. n-factor is defined as the number of hydroxyl ions replaced by 1 mole of base in a reaction.