Question

Number of atoms of iron present in 100 g Fe2O3 having 20% purity

Answer 1

Given condition ⇒
Mass of Fe₂O₃ = 100 g.
Mass of Pure Fe₂O₃ = 20% of 100
 = 20 g.

Molar Mass of Fe₂O₃ = 160 g/mole.

No. of moles of Fe₂O₃ = 100/160
 = 0.625 moles.

Now, 1 mole of Fe₂O₃ contains 2 × 56 atoms of Fe.
∴ 0.625 mole of Fe₂O₃ contains 112 × 0.625 = 70 g of Fe.


No. of Moles of Fe = 70/56
 = 1.25 moles.

No. of atoms of Fe = 1.25 × 6.022 × 10²³
 = 7.5275 × 10²³ atoms. 

Now, this much atoms is present in 100 g of Fe₂O₃.
Now, Purity of Ferric Oxide is also given, so we can find the number of atoms of Iron in Pure Ferric Oxide. 

∴ No. of moles of Pure Fe₂O₃ = 20/160
   = 0.125 moles.

∵ Now, 1 mole of Fe₂O₃ contains 2 × 56 atoms of Fe.
∴ 0.125 moles of Fe₂O₃ contains 112 × 0.125 =  14 g of Fe.

∴ No. of atoms of Fe = 14/56 × 6.022 × 10²³
 = 1.5055 × 10²³ atoms.

Hence, the number of the Electrons of Fe is pure Fe₂O₃ is 1.5055 × 10²³.


Hope it helps.

Answer 2

Answer:

0.25 x 6.022 x 10^23