Q11 of 26 Page 1

An element crystallizes in a f.c.c. lattice with a cell edge of 400 pm. The density of the element is 7g cm-3. How many atoms are present in 280 g of the element?

As it is face lattice,


No of atoms per unit cell (z) =4


Edge length (a)=400pm = 400*10-10cm


Density (d)=7g/cm3


Mass =280g


No .of atoms =?


Volume of cell V=a3=(400*10−10)3=64*10−24cm3


Volume of 280 g of element =m/d


=280/7


=40cm3


No .of unit in 40cm3=


=


=


=6.25×1023unitcells


In FCC each unit cell contain 4 atoms .


Hence total number of atoms in 6.25×1023


units cells are


=4×6.25×1023


=25×1023atoms


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