a mixed oxide of calcium, an unknown metal and oxygen has cubic crystals. in a primitive cubic unit cell of calcium ions the unknown metal ( at wt -48) ions is present at the centre and oxide ions are present in some lattice points the edge length of the cube is 500 pm and density is 1.8g/c . find the oxidation state of metal ion
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Answer:
The length of edge of the unit cell is 5 A
o
or 5×10
−8
cm
The volume of the unit cell is (5×10
−8
cm)
3
=1.25×10
−22
cm
3
The density is 4.0 g cm
−3
The mass of the unit cell is 4.0 g cm
−3
×1.25×10
−22
cm
3
=5.0×10
−22
g......(1)
The mass of one mole of FeO is 55.8+16=71.8 g.
The mass of one molecule of FeO will be
6.023×10
23
71.8
=1.192×10
−22
g.....(2)
Divide (1) with (2) to get number of FeO formula units per unit cell.
1.192×10
−22
5.0×10
−22
g
≃4
Then the number of Fe
2+
and O
2−
ions
present in each unit cell will be four Fe
2+
and four O
2−
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