
Ga (atomic mass \[70\] u) crystallises in a hexagonal close packed structure. The total number of voids in \[0.581{\text{ }}g\] of Ga is ________\[ \times {\text{ }}{10^{21}}\]. (Round off to the nearest integer). [Given:\[{N_A}\; = {\text{ }}6.023{\text{ }} \times {\text{ }}{10^{23}}\]]
Answer
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Hint: Arrangement of atoms in solid phase repeated over three dimensions leads to formation of two types of closed packed structure. One is hexagonal closed packing and other is cubic closed packing. ABA type of arrangement is observed in case of hexagonal closed packing (HCP). The second layer is in the void of layer A and the third layer is again at the same position as layer A. This leads to formation of tetrahedral and octahedral voids.
Complete Step by Step Solution:
Given data: atomic mass of Ga\[ = {\text{ }}70u\]
Given mass of Ga\[ = {\text{ }}0.581g\]
Therefore, number of moles of Ga \[ = \frac{{given{\text{ }}mass}}{{atomic{\text{ }}mass}}\]
\[ = \frac{{0.581}}{{70}}\]
1 mole of an atom contains Avogadro\[{N_A}\; = {\text{ }}6.023{\text{ }} \times {\text{ }}{10^{23}}\] number of atoms
Therefore, \[\frac{{0.581}}{{70}}\] moles of Ga contains $ = 6.023{(10)^{23}}\frac{{0.581}}{{70}}$number of Ga atoms
In hexagonal close packing, n number of octahedral and 2n number of tetrahedral voids are observed. Hence for a single atom, there are\[\;2\]tetrahedral and \[1\]octahedral void.
Thus, total number of void in 1 atom\[ = 3\]
Total number of voids $6.023{(10)^{23}}\frac{{0.581}}{{70}}$ number of Ga atoms = $(3)6.023{(10)^{23}}\frac{{0.581}}{{70}}$
\[ = {\text{ }}15{\left( {10} \right)^{21}}\]
Thus, the correct integer Value is $15$.
Note: In a unit cell of hexagonal close packing, no. of atoms, \[Z = 6\]. Hence a Unit cell has 6 octahedral voids and \[12\] tetrahedral voids. The tetrahedral voids are positioned at body diagonal of the cell and the octahedral voids are positioned at body centre and at the edge centre. Thus, at body diagonal overall there are \[\;1\]octahedral voids and \[2\] tetrahedral voids found.
Complete Step by Step Solution:
Given data: atomic mass of Ga\[ = {\text{ }}70u\]
Given mass of Ga\[ = {\text{ }}0.581g\]
Therefore, number of moles of Ga \[ = \frac{{given{\text{ }}mass}}{{atomic{\text{ }}mass}}\]
\[ = \frac{{0.581}}{{70}}\]
1 mole of an atom contains Avogadro\[{N_A}\; = {\text{ }}6.023{\text{ }} \times {\text{ }}{10^{23}}\] number of atoms
Therefore, \[\frac{{0.581}}{{70}}\] moles of Ga contains $ = 6.023{(10)^{23}}\frac{{0.581}}{{70}}$number of Ga atoms
In hexagonal close packing, n number of octahedral and 2n number of tetrahedral voids are observed. Hence for a single atom, there are\[\;2\]tetrahedral and \[1\]octahedral void.
Thus, total number of void in 1 atom\[ = 3\]
Total number of voids $6.023{(10)^{23}}\frac{{0.581}}{{70}}$ number of Ga atoms = $(3)6.023{(10)^{23}}\frac{{0.581}}{{70}}$
\[ = {\text{ }}15{\left( {10} \right)^{21}}\]
Thus, the correct integer Value is $15$.
Note: In a unit cell of hexagonal close packing, no. of atoms, \[Z = 6\]. Hence a Unit cell has 6 octahedral voids and \[12\] tetrahedral voids. The tetrahedral voids are positioned at body diagonal of the cell and the octahedral voids are positioned at body centre and at the edge centre. Thus, at body diagonal overall there are \[\;1\]octahedral voids and \[2\] tetrahedral voids found.
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