Question

Calculate equivalent mass of \[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O\]
A. \[{\text{126}}\,{\text{g}}\]
B. \[{\text{118}}{\text{.5}}\,{\text{g}}\]
C. \[{\text{102}}{\text{.4}}\,{\text{g}}\]
D. \[{\text{110}}\,{\text{g}}\]

Answer 1

Hint: The molar mass of a compound is calculated by adding the molar masses of all the constituent atoms. It is the mass of an element divided by the amount of the given substance.

Complete step by step answer:
We know that the molar mass of a compound is the sum of the molar mass of all the elements involved in the compound.
We will now calculate the molar mass of \[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O\].
Molar mass of \[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O = \,948.7\,{\text{gmo}}{{\text{l}}^{ - 1}}\]
We know that number of electrons involved = 8
Now we will calculate the equivalent weight of \[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O\]
\[\therefore {\text{Equivalent}}\,{\text{weight}}\, = \,\dfrac{{948.7}}{8} = \,118.5\,{\text{g}}\]
Therefore, the required equivalent mass of \[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O\] is \[\,118.5\,{\text{g}}\].

So, the correct answer is Option B.

Additional information:
Molecular weight is also known as molecular mass. It is the mass of a substance which is based on the atomic weight of 12- carbon which is 12. Molecular weight is the total weight after summing up the atomic weights of all the atoms present in the molecular formula of a compound. For example,we can calculate the molecular mass of glucose in the following way.

Note: The number of atoms in 1 mole is equal to the Avogadro’s number denoted as \[\left( {{{\text{N}}_{\text{A}}}} \right)\]. The value of Avogadro’s number is \[6.022 \times {10^{23}}\]. The molar mass is given as mass divided by mole that is,
Molar mass = mass/mole = g/mole. The mole concept also concludes that the atomic mass of the Carbon-12 is equivalent to the 12 atomic mass units. Also, the mass of Carbon-12 is exactly \[{\text{12 grams}}\] per mole.