Question

Which of the statements are true? (Mw is the molecular weight of the respective compounds)
This question has multiple correct options
A.The equivalent weight of $C{a_3}{\left( {P{O_4}} \right)_2}$is $\dfrac{{{\text{Mw}}}}{6}$
B.The equivalent weight of $N{a_3}P{O_4}.12{H_2}O$is $\dfrac{{{\text{Mw}}}}{3}$
C.The equivalent weight of$\;{K_2}S{O_4}$ is $\dfrac{{{\text{Mw}}}}{2}$
D.The equivalent weight of potash alum ${K_2}S{O_4}A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O$is $\dfrac{{{\text{Mw}}}}{8}$

Answer 1

Hint : We have to calculate the valence factor for each molecule and then put it into the equivalent weight formula (student don’t need to calculate the molecular weight)
Formula Used: ${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight (Mw)}}}}{{{\text{Valency factor}}}}$

Complete step by step solution :
Let’s start with the concept of valency factor; valency factor is the number of electrons gained or lost by an element to gain its nearest inert gas configuration. We can also say that it is the number of electrons gained or lost in a chemical compound.
So now we know about the valency factor so let’s calculate the valency factor for each molecule present in the question.
Case 1
Valency factor for$\;C{a_3}{\left( {P{O_4}} \right)_2}$, so as we can see there are 3 Ca and 2 PO4’s in this molecule. So, the valency of Ca is +2 and since there are 3 of it this means the total valency of 3 Ca is 6. Similarly, in case of PO4 the valency is -3 and since there are 2 of them which means the total valency of PO4 is 6. So the valency factor for $\;C{a_3}{\left( {P{O_4}} \right)_2}$is 6.
Case 2
Valency factor for$N{a_3}P{O_4}.12{H_2}O$, so we can see that this molecule is a hydrated molecule and there are 3 Na and one PO4. So, the valency of Na is +1 and since there are 3 of them this means the total valency is 3. Similarly, in the case of PO4 the valency of PO4 is -3. So, the valency factor of $N{a_3}P{O_4}.12{H_2}O$is 3.
Case 3
Valency factor for${K_2}S{O_4}$, we can see it consists of 2 K and one SO4. So, the valency of K is +1 and since there are 2 of them then the total valency is +2. Similarly, the total valency of SO4 is -2. Hence, the valency factor of ${K_2}S{O_4}$ is 2.
Case 4
Valency factor for${K_2}S{O_4}A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O$ just like case 2 this one also is a hydrated product and consists of 2 K, 4 SO4 and 2Al. The valency of K is +1 and since there are 2 K, this means total valency is +2. For SO2 the valency is -2 and since there are 4 this means that the total is -8. For Al the valency is +3 and since there are 2 Al the total is +6. So, the positive total valency is +8 and negative is -8. So, the valency factor is 8.
Now we know that the equivalent weight is given using formula, ${\text{Equivalent weight = }}\dfrac{{{\text{Molecular weight (Mw)}}}}{{{\text{Valency factor}}}}$
So for $\;C{a_3}{\left( {P{O_4}} \right)_2}$ the equivalent weight will be $\dfrac{{{\text{Mw}}}}{6}$
For $N{a_3}P{O_4}.12{H_2}O$ the equivalent weight will be $\dfrac{{{\text{Mw}}}}{3}$
For ${K_2}S{O_4}$ the equivalent weight will be $\dfrac{{{\text{Mw}}}}{2}$
For ${K_2}S{O_4}A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O$ the equivalent weight will be $\dfrac{{{\text{Mw}}}}{8}$
Therefore, we can conclude all options of this question are correct A., B., C., and D.

Note : We must know that the equivalent weight is used in many questions and hence, we should know how to calculate the equivalent weight.
We must know that the equivalent weight is the mass of one equivalent, which is the mass of a given substance which will combine with or displace a fixed quantity of another substance.
The questions such as Normality, atomic mass calculation, etc. require the concept of equivalent weight. So, students should diligently practice this concept.