Q15 of 46 Page 1

An element crystalizes in FCC lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. Calculate the number of atoms in 108 g of the element.

Given:


Edge length = 300pm = 300×10-8 cm


Density = 10.8gcm-3


To find: number of atoms in 108 g of the element.


Solution:


Volume of a unit cell = (edge length)3


= (300×10-8 cm)3


= 2.7×10-23 cm3


Volume of the given 108 g of unit cell



The number of unit cells is



For one FCC unit cell the number of atoms are 4, thus for the given weight of substance, the number of atoms is,




There are present in 108 g of the given substance.


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