1.

Iron has a body-centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm^(-3). Use this information to calculate Avogadro's number. Atomic mass of Fe = 56.0 u.

Answer»

Solution :Given `a=286.65" PM "=286.65xx10^(-10)cm, d=7.87gcm^(-3), z=2" (for bcc)"`
USE the relation
`d=(zM)/(a^(3)N_(A)) or N_(A)=(zM)/(a^(3)d)`
Substituting the VALUES in the above equation, we have
`N_(A)=(2xx56)/((286.65xx10^(-10))^(3)xx7.87) or N_(A)=6.02xx10^(23)`


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