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Concentration Terms - An Introduction (w/w etc)

0.804 g sampl...

Question

$0.804$ g sample of iron ore was dissolved in acid. Iron was oxidised to $+ 2$ state and it required $47.2$ mL of $0.112 N K M n O_{4}$ solution for titration.
If the percentage of $F e$ in the ore (as nearest integer value) is $6 X + 1$ . Then $X$ is:

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Solution

The redox changes are as follows:

$F e ⟶ F e^{2 +} + 2 e^{-}$
$F e^{2 +} ⟶ F e^{3 +} + e^{-}$
$5 e + M n^{7 +} ⟶ M n^{2 +}$

Meq. of $F e^{2 +}$ $=$ Meq. of $K M n O_{4}$

$\frac{w}{(\frac{56}{1})} \times = 47.2 \times 0.112$ $⟹ w_{F e^{2 +}} = 0.296$ g

$∴$ Purity of $F e = \frac{0.296 \times 100}{0.804} = 36.82 %$

Now, $F e_{3} O_{4} ⟶ 3 F e$
$∵ 3 \times 56$ g $F e$ is obtained by $232$ g $F e_{3} O_{4}$

$∴ 0.296$ g $F e$ is obtained by $\frac{232 \times 0.296}{56 \times 3} = 0.409$ g $F e_{3} O_{4}$
$∴ %$ of $F e_{3} O_{4} = \frac{0.409}{0.804} \times 100 = 50.84 %$

$%$ of $F e = 36.82 %$ .
Thus, $6 X + 1 + 36.82$
then, $X$ = $6$

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