Question

The equivalent mass of ${{K}_{2}}S{{O}_{4}}A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.12{{H}_{2}}O$ is (M=Molar mass of salt):
(1)$\dfrac{M}{12}$
(2) $\dfrac{M}{10}$
(3) $\dfrac{M}{8}$
(4) $\dfrac{M}{6}$

Answer 1

Hint: The answer here includes the basic concept of chemistry that deals with the calculation of equivalent mass by using the formula which is given by the ratio of molar mass of salt to that of its valency.

Complete step – by - step solution:
We are very familiar with the concept of general chemistry that deals with the various calculations used for the determination of mass of compounds which includes equivalent weight, molecular weight, molar mass etc.
Now, we shall see the calculation of equivalent weight of the given salt which will give us the required answer.
First we shall go through the definition of equivalent weight based on which we can deduce the answer.
- Equivalent weight is defined as the mass of one equivalent that is nothing but the mass of a given substance which will combine with or also displaces the fixed quantity of another substance.
- Equivalent mass is thus given by the ratio of molar mass of the compound to that of its valency and thus can be written in the formula as,
$Eq.Wt.=\dfrac{M}{valency}$ ………(1)
where M is the molar mass.
Now, let us check the valency of the salt and since water does not add to equivalent weight, the equivalent weight of the salt will be considered and is as follows,
valence of potassium ion = +1
For aluminium = +3
for the sulphate ion = _2
Thus the total valency = 1+3+2 =6
Therefore equivalent weight from equation (1) is M/6

Thus, the correct answer is option 4) $\dfrac{M}{6}$

Note: Do not be confused about equivalent weight and molecular weight. Molecular weight is the mass of the substance which is based on the one – twelfth of carbon – 12 atoms and equivalent weight is the proportional mass of chemical entities which displace or combine with other entities.