Question

The mass percentage of ${Fe}^{+3}$ ion present in $F{e}_{0.93}{O}_{1.0}$ is?

1)15%.

2) 5.5%.

3) 10%.

4) 11.5%

Answer 1

Step 1:

Let x atoms of $F{e}^{3+}$ ions be present.

This means x no. of $F{e}^{3+}$ ions have been replaced by $F{e}^{2+}$ ions:

Thus, no. of $F{e}^{2+}$ ions = 0.93−x

For electrical neutrality,

2 (0.93−x) + 3x - 2 = 0

or, 1.86 + x = 2

or, x = 0.14

Step 2:

Fraction of $F{e}^{3+}$ = 0.14, $F{e}^{2+}$ = 0.93−0.14 = 0.79

Formula : $F{e}_{0.79}^{2+}F{e}_{0.14}^{3+}{O}_{1.0}^{2-}$

Therefore, total molar mass = $0.93\times 56+1\times 16=68.08$

% of Fe(III) = $\frac{0.14\times 56}{68.08}\times 100=11.5\%$

Hence, option (4) is correct.