Question

What is the equivalent weight of $ A{l_2}{(S{O_4})_3} $ ?

Answer 1

Hint: Equivalent weight of a substance is defined as the weight of the substance that completely reacts or displaces the fixed quantity of other substances. An element’s equivalent weight can also be given as the amount that displaces or combines with 1.008 g of Hydrogen or 8.0g of Oxygen or 35.5g of Chlorine.

Complete answer:
The equivalent weight of any compound can be given by the formula: $ \dfrac{{Molecular{\text{ }}Mass}}{{Valency{\text{ }}Factor}} $ --(1)
First, we will find the molecular mass of $ A{l_2}{(S{O_4})_3} $ . Aluminium Sulphate is a salt that consists of Aluminium, Sulphur and Oxygen. One mole of aluminium sulphate consists of 2 aluminium atoms, 3 sulphur atoms and 12 oxygen atoms. The molar mass of $ A{l_2}{(S{O_4})_3} $ will be equal to the molar masses of the elements that make up the compound. The molar masses of the respective elements are: $ {M_{Al}} = 27g/mol,{M_S} = 32g/mol,{M_O} = 16g/mol $
The total molar mass of aluminium sulphate will be: $ 2 \times {M_{Al}} + 3 \times {M_S} + 12 \times {M_O} $
 $ 2 \times 27 + 3 \times 32 + 12 \times 16 = 342g/mol $
The valency factor can be found out by disintegration of the Aluminium Sulphate salt. The reaction is:
 $ A{l_2}{(S{O_4})_3} \to 2A{l^{ + 3}} + 3SO_4^{2 - } $
The valency factor is equal to the total charge present in the compound which is equal to $ 2 \times {( + 3)_{Al}} = 3 \times {( - 2)_{S{O_4}}} = | - 6| = 6 $
The equivalent weight of $ A{l_2}{(S{O_4})_3} $ $ = \dfrac{{342}}{6} = 57g/mol $ (from equation (1))
Hence the equivalent weight of $ A{l_2}{(S{O_4})_3} $ is 57 g/mol.

Note:
The equivalent weight of various acids, bases and salts can be given by the formulas:
Equivalent weight of an acid = molecular mass of the acid/basicity
Equivalent weight of base = Molecular mass of base/acidity
Equivalent weight of salt = Molecular mass of the salt / Total valency of the compound.
The equivalent mass of a compound is mainly used to determine the normality of any solution. The formula for normality is: $ Normality = \dfrac{{Mass{\text{ }}of{\text{ }}solute}}{{equivalent{\text{ }}weight \times volume{\text{ }}of{\text{ }}solution}} $ .