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The percentage of oxygen in $F e_{2} (S O_{4})_{3}$ is

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46 %

50 %

52 %

48 %

answer is D.

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Detailed Solution

The percentage of oxygen in

F e_{2} (S O_{4})_{3}

48 %

.
Atomic mass of,

Fe = 56 u

Atomic mass of

S = 32 u

Atomic mass of oxygen

= 16 u

Molecular mass of

F e_{2} (SO)_{3} = (56 \times 2) + (32 \times 3) + (16 \times 12)

\Rightarrow = 400 u

Total mass of oxygen

= 16 \times 4 \times 3

\Rightarrow = 192 g

So percentage of oxygen in

F e_{2} (S O_{4})_{3} = \frac{192}{400} \times 100

\Rightarrow 48 %

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