Question

In the following conversion of sulphide of phosphorous \[{P_4}{S_3} \to {P_2}{O_5} + S{O_2}\]
Equivalent weight of\[{P_4}{S_3}\] (molecular weight = M) is:
A.\[\dfrac{M}{{18}}\]
B.\[\dfrac{M}{{14}}\]
C.\[\dfrac{M}{{32}}\]
D.\[\dfrac{M}{{38}}\]

Answer 1

Hint:We should be aware of the fact that gram equivalent and equivalent weight are the same terms. The concepts of calculating valency for a particular element, balancing the redox reactions should be clear to us.

Formula used:
Equivalent weight= Atomic weight/valency
Equivalent weight of a compound = Molecular weight/n-factor.

Complete step by step answer:
In general, equivalent weight is calculated as atomic weight of a substance divided by its valency, or usual valance.
In layman language, it is the mass of the given substance ’A’ that will combine or displace a fixed quantity ‘B’ from another substance.
But, in the given question, we observe that the question deals with molecular weight and equivalent weight of a compound.
So, let us discuss about relation between equivalent weight and molecular weight.
Equivalent weight of a compound = Molecular weight/n-factor.
n-factor is defined as total positive or negative charge when a compound liquidates or dissolutes.
In the given reaction,
 \[{P_4}{S_3} \to {P_2}{O_5} + S{O_2}\]
Balancing the reaction,
We get
\[{P_4}{S_3} + 16{H_2}O \to 2{P_2}{O_5} + 3S{O_2} + 32{H^ + } + 32{e^ - }\]
Therefore, n-factor= 32
Thus Equivalent weight = \[\dfrac{M}{{32}}\]

Thus, option C. is the correct option for the given question.

Additional Information:
The concept of equivalent weights is widely used in volumetric and gravimetric analysis.
It also depends upon the number of moles of the compound in the reaction.
Thus, we can conclude that equivalent weight varies from reaction to reaction.

Note: In the field of polymers, reactivity of polymer is inverse of the equivalent weight.
Equivalent weight has the dimensions as that of mass, whereas atomic weight is dimensionless.