In $\mathrm{Fe}_{0.96} \mathrm{O}, \mathrm{Fe}$ is present in +2 and +3 oxidation state, what is the mole…
In $\mathrm{Fe}_{0.96} \mathrm{O}, \mathrm{Fe}$ is present in +2 and +3 oxidation state, what is the mole fraction of $\mathrm{Fe}^{2+}$ in the compound?
12/25
25/12
1/12
11/12
Solution
Mole fraction is unit of amount of constituent, divided by total amount of all constituents in a mixture.
Let us consider, there are ' $x$ ' molecular fraction of $\mathrm{Fe}^{2+}$ and $(0.96-x)$ molecular fraction of $\mathrm{Fe}^{3+}$.
Using charge conservation formula, We get,
$
\begin{aligned}
x(+2)+(0.96-x)(+3)-2 & =0 \\
2 x-3 x+2.88-2 & =0 \\
x & =0.88
\end{aligned}
$
$\therefore$ Fraction of $\mathrm{Fe}^{2+}=0.88$
and fraction of $\mathrm{Fe}^{3+}=0.96-0.88=0.08$
Percentage of $\mathrm{Fe}^{2+}=\frac{0.88}{0.96} \times 100=91.6 \%$
Mole fraction of $\mathrm{Fe}^{2+}=\frac{0.88}{0.96}=\frac{11}{12}$
Hence, mole fraction of $\mathrm{Fe}^{2+}$ is $\frac{11}{12}$