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Iron has body centred cubic cell with a ...

Iron has body centred cubic cell with a cell edge of 286.5 pm. The density of iron is 7.87 g `cm^(-3)`. Use this information to calculate Avogadro's number. (Atomic mass of Fe = 56 `mol^(-3)`)

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Here , a = 286.65 pm = `286.65 xx 10^(-10)` cm
` d = 7.87 g cm^(-3)`
M = 56
Z = 2 , Na ?
Using the formula `" " d = (Z.M)/(a^(3).N_(A)) implies N_(A) = (Z.M)/(a^(3).d)`
` therefore " " N_(A) = (2 xx 56)/((286.65 xx 10^(-10))^(3) xx 7.87)`
N` = (112)/((2.87 xx 10^(-8))^(3) xx 7.87) = (112)/(23.63 xx 7.87 xx 10^(-24))`
`= (112)/(185.97 xx 10^(-24)) = 0.6022 xx 10^(24) = 6.022 xx 10^(23)`
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