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Total number of volds present in 38 mg of osmium metal which crystallizes as hcp lattice is? (Atomic mass of Os is 190 amu, NA = 6 x 102³)

Total number of volds present in 38 mg of osmium metal which crystallizes as hcp lattice is? (Atomic mass of Os is 190 amu, NA = 6 x 102³)

Grade:12

2 Answers

Priyanshu Gujjar
94 Points
2 years ago

The total number of octahedral void(s) per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 41th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.
12 octahedral voids located at each edge and shared between four unit cells=12×41=3
Total number of octahedral voids =4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to 4/4=1.

Anish
105 Points
2 years ago
askiitians.com
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 Forum Physical Chemistry Total number of volds present in 38 mg of...
Total number of volds present in 38 mg of osmium metal which crystallizes as hcp lattice is? (Atomic mass of Os is 190 amu, NA = 6 x 102³)
 
kiriti, 10 days ago
Grade:12
 1 Answers
Priyanshu Gujjar 
 101 Points
3 days ago
The total number of octahedral void(s) per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only  
4
1
 th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.
12 octahedral voids located at each edge and shared between four unit cells=12× 
4
1
 =3
Total number of octahedral voids =4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to 4/4=1.

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