Question

What weight of $FeS{O_4}{\text{ }}\left( {molecular{\text{ weight = 152}}} \right)$ will be oxidised by $200{\text{ ml}}$ of normal $KMn{O_4}$ solution in acidic solution.
$(i){\text{ 30}}{\text{.4 g}}$
$(ii){\text{ 60}}{\text{.8 g}}$
$(iii){\text{ 121}}{\text{.6 g}}$
$(iv){\text{ 15}}{\text{.8 g}}$

Answer 1

The number of equivalents of $FeS{O_4}$ will be equal to the number of equivalents of $KMn{O_4}$ .
Then we will calculate the n-factor of each compound. With the help of the molarity equation we will find the number of moles of $FeS{O_4}$ . Thus we can find the weight of $FeS{O_4}$ .
${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{M}}_2}{{\text{V}}_2}$

Complete answer:
Since the number of equivalents of $FeS{O_4}$ is equal to the number of equivalents of $KMn{O_4}$ , we can find the number of moles of $FeS{O_4}$ . Firstly we find out the $n -$ factor of the $KMn{O_4}$ .
It can be calculated by using the below reduction process,
$M{n^{ + 7}}{\text{ }} \to {\text{ M}}{{\text{n}}^{ + 2}}$
$n -$ factor is the change of oxidation state for a particular atom. For manganese $n -$ factor is $5$ .
Let ${{\text{n}}_{\text{1}}}$ be the number of moles of $KMn{O_4}$ and ${\text{n}}{{\text{f}}_1}$ be its $n -$ factor. Similarly ${{\text{n}}_{\text{2}}}$ be the number of moles of $FeS{O_4}$ and ${\text{n}}{{\text{f}}_{\text{2}}}$ be its $n -$ factor.
Then we can write that,
${{\text{n}}_{\text{1}}}{\text{ x n}}{{\text{f}}_1}{\text{ = }}{{\text{n}}_2}{\text{ }} \times {\text{ n}}{{\text{f}}_2}$
Also we know that, ${\text{n = M }} \times {\text{ V}}$ . Therefore we can write that,
${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{n}}_2}{\text{ }} \times {\text{ n}}{{\text{f}}_2}$
Also $n -$ factor for $FeS{O_4}$ is equal to $1$ .
${\text{1 }} \times {\text{ }}\dfrac{{200}}{{1000}}{\text{ = }}{{\text{n}}_2}{\text{ }} \times {\text{ 1}}$
Therefore, ${{\text{n}}_{\text{2}}} = {\text{ 0}}{\text{.2}}$
The molecular weight of $FeS{O_4}$ is given as $152{\text{ g}}$ . The mass of $0.2{\text{ mole }}$ of $FeS{O_4}$ will be $152{\text{ }} \times {\text{ 0}}{\text{.2 = 30}}{\text{.4 g}}$ .
Thus we can say that the amount of $FeS{O_4}$ oxidised will be equal to $30.4{\text{ g}}$ by the given volume of $KMn{O_4}$ . Thus the correct answer is option $(i){\text{ 30}}{\text{.4 g}}$ .

Note:
The value of $n -$ factor can be obtained by the oxidation and reduction reaction. It is the change of oxidation state of the atom in respective compounds. Also the weight of a given compound is found by multiplying the number of moles with the molecular weight of the compound. Iron gets oxidised in the above reaction while manganese gets reduced. Iron changes its oxidation from $F{e^{ + 2}}$ to $F{e^{ + 3}}$ . This is because $n -$ factor is taken as one.