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Ostwald Dilution Law

The ratio of ...

Question

The ratio of $F e^{3 +}$ and $F e^{2 +}$ ions in $F e_{0.9} S_{1.0}$ is

0.28

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0.5

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2

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4

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Solution

The correct option is A $0.28$
$F e^{3 +} = x$
$F e^{2 +} = 0.93 - x$
Using charge balance
$F e_{0.9} S_{1.0}$ =Charge on neutral compound $= 0$
$3 \times x 2 \times (0.93 - x)      + 2 \times 1 = 0$
Charge on $F e_{0.9}$
$(F e^{2 +} & F e^{3 +})$
$\Rightarrow 3 x + 2 (0.9 - x) = 2 \times 2$
$3 x + 1.8 - 2 x = 2$
$x = 2 - 1.8$
$x = 0.2$
Thus the no. of $F e^{+ 2}$ ion $= 0.93 - x$
$= 0.93 - 0.2$
$= 0.73$
Ratio $\frac{F e^{3 +}}{F e^{2 +}} = \frac{0.2}{0.73}$
$= 0.28$

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