Question

$3{\text{ }}mol$of a mixture of $FeS{O_4}$ and $F{e_2}{\left( {S{O_4}} \right)_3}$ required 100ml of $2M{\text{ }}KMn{O_4}$ solution in acidic medium. Hence, mole fraction of $FeS{O_4}$ in the mixture is:
A.$\dfrac{1}{3}$
B.$\dfrac{2}{3}$
C.$\dfrac{2}{5}$
D.$\dfrac{3}{5}$

Answer 1

Hint : We have to calculate mole fraction of$FeS{O_4}$in the mixture. We can understand that $FeS{O_4}$ had Fe in +2 oxidation state and $F{e_2}{\left( {S{O_4}} \right)_3}$has Fe in +3 states. So, $KMn{O_4}$ will oxidise only from $F{e^{2 + }}$to $F{e^{3 + }}$

Complete step by step solution :
We can write the chemical reaction from a given condition.
Therefore,
$2KMn{O_4} + {\text{ }}8{H_2}S{O_4} + {\text{ }}10FeS{O_4} \to {K_2}S{O_4} + {\text{ }}2MnS{O_4} + {\text{ }}5F{e_2}{\left( {S{O_4}} \right)_3} + {\text{ }}8{H_2}0$
In the reaction, the potassium permanganate oxide the $FeS{O_4}$ to$F{e_2}{\left( {S{O_4}} \right)_3}$
Also we know that,
Equivalents of $FeS{O_4}$= Equivalents of $KMn{O_4}$
Or, Equivalents of $FeS{O_4}$= $0.1L \times (2 \times 5)N$
Therefore, Equivalents of $FeS{O_4}$= 1
Or,$Moles{\text{ }}of{\text{ }}FeSO4 = \dfrac{{Equivalent}}{{n - factor}} = \dfrac{1}{1}$ (since$FeS{O_4}$ n=1)
Therefore, Mole fraction of $FeS{O_4}$ in the mixture =$\dfrac{{moles{\text{ }}of{\text{ }}FeS{O_4}}}{{Moles{\text{ }}of{\text{ }}mixture}} = \dfrac{1}{3}$
We can conclude that the mole fraction of $FeS{O_4}$ in mixture =$\dfrac{1}{3}$.

And hence the option A is correct.

Note : The mole fraction is also known as the amount fraction. It is identical to the number fraction, which is defined as the number of molecules of a constituent. We can divide ${N_i}$ by the total number of all molecules${N_{tot}}$. We can use a mole fraction very frequently in the construction of different phase diagrams. It has various advantages:
It does not depend on fluctuation of temperature and also do not require any knowledge of the other densities of the phases involved. In the mole fractions, $x{\text{ }} = {\text{ }}0.1{\text{ }}and{\text{ }}x{\text{ }} = {\text{ }}0.9$, the value of 'solvent' and 'solute' are reversed particularly. In a mixture of ideal gases, the mole fraction can be expressed as the ratio of partial pressure to the total pressure of the mixture.