Calculate the number of unit cells in 8.1 g of aluminum
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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=27g of Aluminium
Also , 1 mole contains Avogadro number of atoms
1mole=6.023×10231mole=6.023×1023 atoms of Aluminium
Therefore
1 mole of Al - 27 g of aluminium - 6.023×10236.023×1023atoms of aluminium
Number of atoms in 8.1g8.1g aluminium is =(8.1×6.023×1023)/27=(8.1×6.023×1023)/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(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=27g of Aluminium
Also , 1 mole contains Avogadro number of atoms
1mole=6.023×10231mole=6.023×1023 atoms of Aluminium
Therefore
1 mole of Al - 27 g of aluminium - 6.023×10236.023×1023atoms of aluminium
Number of atoms in 8.1g8.1g aluminium is =(8.1×6.023×1023)/27=(8.1×6.023×1023)/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(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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2
Explanation:
n = given mass/molar mass
= {8.1}/{27} mol
Number of atoms = 8.1/27 × 6.022 ×× 10²³
Number of atoms in one unit cell = 4 (fcc)
Number of unit cell = [8.1/27 × 6.022 × 10²³]
FINAL RESULT = 4.5 × 10²²
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