Calculation for the number of unit cells (using log table)

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) (CBSE 2017)

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.

 

For more questions, visit - Practise Problems on 'The number of atoms in a unit cell'- Solid State

 

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