Question

The equivalent mass of $N{{a}_{2}}HP{{O}_{4}}$ can be:
A. $\dfrac{M}{2}$ as base
B. $\dfrac{M}{1}$ as acid
C. $\dfrac{M}{1}$ during the change, $2N{{a}_{2}}HP{{O}_{4}}\to N{{a}_{4}}{{P}_{2}}{{O}_{7}}+{{H}_{2}}O$
D. either of these

Answer 1

Hint: Equivalent mass is the mass of a given substance which will react or displace a fixed quantity of another substance. It can be calculated by
Eq. mass = $\dfrac{Atomic~number}{valency}$
However, if there is a complex compound or a salt. Then there are other methods to calculate the equivalent mass too. In the above question, we will take n-factor over this formula because it’s a compound.

Complete step by step solution :
In this question, we will calculate with the help of Equivalent mass= molar mass/ n-factor, we will look at all the options given here:
A. In this option equivalent mass of $N{{a}_{2}}HP{{O}_{4}}$ will be $\dfrac{M}{2}$ as a base when it will react with 2 moles of monobasic acid.
B. in this option, equivalent mass of $N{{a}_{2}}HP{{O}_{4}}$ will be $\dfrac{M}{1}$ as acid because it has one replaceable Hydrogen atom in its formula, which will act as n-factor.
C. $2N{{a}_{2}}HP{{O}_{4}}\to N{{a}_{4}}{{P}_{2}}{{O}_{7}}+{{H}_{2}}O$
- In this reaction, it is giving one replaceable H-atom, therefore its n-factor will be 1 and hence, the equivalent ma becomes $\dfrac{M}{1}$.
All the options given above are true, therefore, the answer to this question is D. Either of these, as equivalent mass can change depending on the conditions.
So, the correct answer is “Option A”.

Note: In this question, we have calculated the equivalent mass by calculating the n-factor or the replaceable amount of Hydrogen atoms, as well as the M represents the molecular mass, in the case of acids and bases. The eq. mass can be found out with the help of other methods such as Oxide method for the oxides.