Question

If M is the molar mass of $P_4$ in the following reaction:
$P_4+3O{H}^{-}+3H_{2}O\to 3H_{2}P{O}_4+P{H}_3$
The equivalent mass of $P_4$ is
A.$M$
B.${\displaystyle \frac{M}{3}}$
C.${\displaystyle \frac{M}{4}}$
D.${\displaystyle \frac{M}{12}}$

Answer 1

Hint: The equivalent mass of an element or a compound is said to be the total number of gram equivalents of the substance. Equivalent mass is given by the ratio between the molar mass of the substance and the n- factor of the substance. To find the equivalent mass of $P_4$ in the given reaction, we must find its n- factor

Complete step by step solution:
The n factor is known as the valence factor and its value varies depending upon the compound under consideration. For instance, for an acid, the n factor is the basicity of the acid and the n factor of a base is its acidity. Generally, the valence factor for a substance is given by the change in its oxidation state.
So the equivalent mass of $P_4$ will be given by the ratio between the molar mass and the change in oxidation state of phosphorus in the reaction.
In $P_4$, the oxidation number of phosphorus is $=0$.
After the reaction, it forms $P{H}_3$ and $H_{2}P{O}_4$. The oxidation state of phosphorus in $P{H}_3$ is $-3$ and that in $H_{2}P{O}_4$ is $+3$. So the valence factor will be equal to $3$.
The equivalent mass of $P_4={\displaystyle \frac{M}{3}}$.

Thus, the correct answer is B.

Note:
Stoichiometry is based upon the very basic laws of chemistry that help to understand it better, namely, the law of conservation of mass, the law of definite proportions (the law of constant composition), the law of reciprocal proportions and the law of multiple proportions . In general, different chemicals combine in definite ratios in chemical reactions. Since matter can neither be created nor destroyed, thus the amount of each element must be the same throughout the entire reaction.