Answer :

Given- fcc lattice, density = 10.8 g cm-3 = β, edge length = 300pm = 300 × 10-12 m = 300 x 10-10cm = a

Find- no. Of atoms in 108g of element


Using the formula for density,


Being fcc lattice Z = 4


Putting all the values on equation we get, 10.8 =


M= 175.6 g


In 175.6 g, 6.022 X 1023 atoms are present


Therefore, in 108g, no. Of atoms will be = 0.61 x 1023 atoms


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