1.

Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm^(-3). Use this information to calculate Avogadro's number. [At. mass of Fe = 56 g mol^(-1)]

Answer»

Solution :APPLY the following relation :
`N_(A)=(zxxM)/(dxxa^(3)`
As it has a BCC structure z = 2.
SUBSTITUTING the values in the above relation, we have
`N_(A)=(2xx56)/((286.65)^(3)xx10^(-30)xx7.87)=6xx10^(23)`.


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