Question: Calculate the number of unit cells in 8.1 g of Aluminium if it crystallizes in a face cubic structure (fcc). (Atomic mass of Al = 27 g/mol)
Solution: The weight of one mole of a substance is equal to its atomic mass in grams, therefore,
1 mole of Al = 27 g of Aluminium
Also, 1 mole contains Avogadro number of atoms
1 mole = 6.023 x 1023 atoms of Aluminium
Therefore,
1 mole of Al = 27 g of Aluminium = 6.023 x 1023 atoms of Aluminium
Nos of atoms in 8.1 g of Aluminium is = (8.1 x 6.023 x 1023) / 27 (Just cross multiply !!)
The face-centered cubic structure contains 4 atoms in one unit cell. Therefore, the number of unit cells that has (8.1 x 6.023 x 1023) / 27 atoms of Aluminium can be obtained as,
(8.1 x 6.023 x 10 23) / (27 x 4) = 4.516 x 1022
Solving the numerical problem using the log table, the rules applied are -
Log (m x n) = Log m + Log n
Log (m/n) = Log m - Log n
Log (mn)= n Log m
From the log table, the values obtained are as follows -
Log 8.1 = 0.9085
Log (6.023 x 1023) = Log 6.023 + Log 1023 = Log 6.023 + 23 Log 10= 0.7798 + 23
Log 27 = 1.4314
Log 4 = 0.6021
Log x = Log (8.1 x 6.023 x 10 23) / (27 x 4) = (0.9085 + 0.7798 + 23) – (1.4314 + 0.6021)
= (1.6883 + 23) – (2.0335)
Separating the characteristic and the mantissa,
= (1 + 0.6883) + 23 - (2 + 0.0335) = 1 + 0.6883 + 23 - 2 - 0.0335
= (1 + 23 -2) + (0.6883- 0.0335) = (22 + 0.6548) = 22.6548
Log x= 22.6548
x= Antilog (22.6548) = 4.516 x 1022
Therefore, 8.1 g of Aluminium contains 4.516 x 1022 unit cells.
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