Q. 15

An element crysta

Answer :

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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