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.