Question

3.92 gm of ferrous ammonium sulphate crystals are dissolved in 100 ml of water, 20ml of this solution requires 18 ml of potassium permanganate during titration for complete oxidation. The weight of $KMn{{O}_{4}}$ present in one litre of the solution is:
(A) 34.76 gm
(B) 12.38 gm
(C) 1.238 gm
(D) 3.476 gm

Answer 1

To solve the given problem, we should know the concepts of normality and the equations for the titration of acid and base based on normality of the solution.
Normality is the number of gram equivalents or mole equivalents of solute present in one litre of a solution.

Complete answer:
Let us solve the given illustration directly, we just need to know the basics of normality as explained below;
Normality-
As we know, normality is the number of gram equivalents or mole equivalents of solute present in one litre of a solution.
Mole equivalent is the number of moles who are actually the reactive units in the reaction. Similar goes with the gram equivalent.
It is expressed as,
$N=M\times n$
where, n is the ratio of molar mass to the equivalent mass of the component.
Normality can simply be expressed as,
$N=\dfrac{{{n}_{g}}}{V}$
where,
V = volume of solution in litres
${{n}_{g}}$ = number of gram equivalents i.e. it is the ratio of given mass to the equivalent mass of a compound.
When we calculate normality during titration (in other words neutralisation), following formula is used,
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
where,
${{N}_{1}}$ = normality of acidic solution
${{V}_{1}}$ = volume of acidic solution
${{N}_{2}}$ = normality of basic solution
${{V}_{2}}$ = volume of basic solution
Illustration-
Given that,
Given weight of $\left( {{\left( N{{H}_{4}} \right)}_{2}}Fe{{\left( S{{O}_{4}} \right)}_{2}}.6{{H}_{2}}O \right)$ = 3.92 gm
Total volume of water = 100 ml
Volume of acidic solution = 20 ml
Volume of basic solution = 18 ml
Now,
Molecular weight of $\left( {{\left( N{{H}_{4}} \right)}_{2}}Fe{{\left( S{{O}_{4}} \right)}_{2}}.6{{H}_{2}}O \right)$ = 392 gm/mol
Molecular weight of $KMn{{O}_{4}}$ = 158 gm/mol
We can describe neutralisation reactions as,
$\begin{aligned} & Mn{{O}_{4}}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O \\\ & F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}} \\\ \end{aligned}$
From above equation we can say,
Equivalent mass of $\left( {{\left( N{{H}_{4}} \right)}_{2}}Fe{{\left( S{{O}_{4}} \right)}_{2}}.6{{H}_{2}}O \right)$ = $\dfrac{392}{1}=392gmmo{{l}^{-1}}e{{q}^{-1}}$
Equivalent mass of $KMn{{O}_{4}}$ = $\dfrac{158}{5}=31.6gmmo{{l}^{-1}}e{{q}^{-1}}$
Thus,
Normality of $\left( {{\left( N{{H}_{4}} \right)}_{2}}Fe{{\left( S{{O}_{4}} \right)}_{2}}.6{{H}_{2}}O \right)$ = $\dfrac{\left( {}^{3.92}/{}_{392} \right)\times 1000}{100}=0.1N$
For titration,
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
so,
$\begin{aligned} & 0.1N\times 20={{N}_{2}}\times 18 \\\ & {{N}_{2}}=\dfrac{1}{9}N \\\ \end{aligned}$
Now, weight of $KMn{{O}_{4}}$ present in 1 L of solution is given as, $\dfrac{1}{9}N\times 31.6gmmo{{l}^{-1}}e{{q}^{-1}}\times 1L=3.476gm$

Therefore, option (D) is correct.

Note:
Do note to use the units properly and do not get confused in concepts of molarity and normality. Though they are interrelated, their multiplying factors differ.