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Standard XII

Physics

Permeability

0.222 g of ir...

Question

$0.222 g$ of iron ore was brought into solution, $F e^{3 +}$ is reduced to $F e^{2 +}$ with $S n C l_{2}$ . The reduced solution required $20 m L$ of $0.1 N K M n O_{4}$ solution. The percentage of iron present in the ore is:

[Equivalent weight of iron is $55.5$ ]

55.5

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45.0

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50.0

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40.0

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Solution

The correct option is C $50.0$ %
Number of equivalents of $F e$ $=$ Number of equivalents of $K M n O_{4}$

$\frac{N V}{1000} = \frac{0.1 \times 20}{1000} = 0.002$

Mass of iron (pure) $= 0.002 \times 55.5 = 0.111 g$

% $p u r i t y = \frac{Mass of pure iron}{Mass of iron ore} \times 100 = \frac{0.111}{0.222} \times 100 = 50$ %.

Hence, option $C$ is correct.

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0.222 g of iron ore was brought into solution, Fe3+ is reduced to Fe2+ with SnCl2. The reduced solution required 20 mL of 0.1 N KMnO4 solution. The percentage of iron present in the ore is: [Equivalent weight of iron is 55.5]

The correct option is C 50.0%Number of equivalents of Fe = Number of equivalents of KMnO4 NV1000=0.1×201000=0.002Mass of iron (pure) =0.002×55.5=0.111 g% purity=Mass of pure ironMass of iron ore×100=0.1110.222×100=50%.Hence, option C is correct.

$0.222 g$ of iron ore was brought into solution, $F e^{3 +}$ is reduced to $F e^{2 +}$ with $S n C l_{2}$ . The reduced solution required $20 m L$ of $0.1 N K M n O_{4}$ solution. The percentage of iron present in the ore is:

[Equivalent weight of iron is $55.5$ ]