Calculate no of unit cells in 8.1 g of al if it crystallise in fcc
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The weight of one mole of a substance is equal to its atomic mass in grams , therefore
1 mole of Al =27g of Aluminium
Also , 1 mole contains Avogadro number of atoms
1mole=6.023×10^23 1mole=6.023×10^23 atoms of Aluminium
Therefore
1 mole of Al - 27 g of aluminium - 6.023×10^23 atoms of aluminium
Number of atoms in 8.1g aluminium is =(8.1×6.023×10^23)/27
The face centered cubic structure contains 4 atoms in one unit cell .
Therefore , the number of unit cells that has (8.1×6.023×1023)/27 atoms of Aluminium can be obtained as ,
(8.1×6.023×1023)/(27×4)=4.516×1022
1 mole of Al =27g of Aluminium
Also , 1 mole contains Avogadro number of atoms
1mole=6.023×10^23 1mole=6.023×10^23 atoms of Aluminium
Therefore
1 mole of Al - 27 g of aluminium - 6.023×10^23 atoms of aluminium
Number of atoms in 8.1g aluminium is =(8.1×6.023×10^23)/27
The face centered cubic structure contains 4 atoms in one unit cell .
Therefore , the number of unit cells that has (8.1×6.023×1023)/27 atoms of Aluminium can be obtained as ,
(8.1×6.023×1023)/(27×4)=4.516×1022
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