Question

$0.84\;g$of a metal carbonate reacts exactly with $40\;ml$ of
$\dfrac{N}{2}{H_2}S{O_4}$. The equivalent weight of the metal carbonates is:
A. $84$
B. $64$
C. $42$
D. $32$

Answer 1

Hint: As we know that, equivalent weight is the amount of an element that reacts with $1$ mole of electron. It is the relative mass of the material which depends on the stoichiometry of the reaction.

Step by step answer: If we let the equivalent weight of metal carbonates as W.
${\rm{Then}}\;{\rm{the}}\;{\rm{number}}\;{\rm{of}}\;{\rm{equivalents}}\;{\rm{metal}}\;{\rm{carbonates}} = \dfrac{{0.84}}{W}$
As we know that, the equivalent weight of any compound is the weight of a compound which undergoes dissociation or association with the other fixed mass of compound.
Now, we will look at the normality of a solution that is basically the number of equivalents of a substance present in $1{\rm{ L}}$ of the solution. We have a relationship between the normality of a solution and molarity or we can say the number of moles and equivalents for a given volume of solution. We can write it as follows:
${\rm{number\; of\; equivalents}} = {\rm{number\;of\; moles}} \times {\rm{equivalent\; factor}}$
Now, this equivalent factor depends on the substance which is under consideration and the reaction that is being carried out.
When we compare the equivalents of metal carbonates and ${H_2}S{O_4}$ we get, this equation shown below. According to question, weight of the metal carbonate and normality of ${H_2}S{O_4}$ is given and we have to find the gram equivalent weight of metal carbonate as follows;
\[
\dfrac{{{\rm{Weight}}\;{\rm{of}}\;{\rm{metal}}\;{\rm{carbonate}}}}{{{\rm{Gram}}\;{\rm{equivalent}}\;{\rm{weight}}}} = N \times V\left( L \right)\\
\Rightarrow \dfrac{{0.84}}{{G.E.M}} = \dfrac{1}{2} \times \dfrac{{40}}{{1000}}\\
\Rightarrow \dfrac{{0.84}}{{G.E.M}} = \dfrac{{20}}{{1000}}\\
\Rightarrow \dfrac{{0.84}}{{G.E.M}} = 0.02
\]
On further simplification,
\[
G.E.M = \dfrac{{0.84}}{{0.02}}\\
\Rightarrow G.E.M = 42\;g
\]

Therefore, the correct answer is C.

Note: Normality is not necessarily better than Molarity. It is just more useful in certain cases, those certain cases being when we talk about acids, bases, salt, pH ionic calculations.