Question

In the given reaction $ {H_2}C{O_3} + NaOH \to NaHC{O_3} + {H_2}O $
Equivalent weight of $ {H_2}C{O_3} $ is ( molar mass of $ {H_2}C{O_3} $ is $ 62g/mol $ )
(A) $ 60 $
(B) $ 32 $
(C) $ 62 $
(D) $ 63 $

Answer 1

Hint: 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 solution
Firstly we will discuss the equivalent weight is the mass of one equivalent that is the mass of a given substance which will combine with or displace the another set of quantities or substance. Further when we talk about the equivalent weight that it has a dimension like the mass unlike atomic weight which is dimensionless. Further 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. Further considering the question we can say that we have
molar mass of $ {H_2}C{O_3} $ is $ 62g/mol $
Moreover, we know that the equivalent weight is equal to the molecular weight divided by the n factor and here we need to know the n factor which is equal to the number of moles of electron lost or gained per molecule. Here in the question the n factor is equal to one so we will calculate the equivalent weight as shown below:
equivalent weight $ = \dfrac{{62}} 1 \\
   = 62 \\ $
here we can calculate the n factor from the equation:
 $ {H_2}C{O_3} + NaOH \to NaHC{O_3} + {H_2}O $
Hence, option C is the correct answer for the given question.

Note
We can say that the n factor can also be dependent on the number of electrons gained or lost.but here as it is simply given in the question we can calculate it by knowing the n factor through the criteria of number of electrons gained or lost.