-->

The Solid State

Question
CBSEENCH12010204

Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro's number (At. Mass of Fe = 55.845 u)

Solution

a = 286.65 pm

a= 286.65 x 10-10cm

Density ( ) = 7.874 g cm-3
At mass of Fe = 55.845 u

Z = 2 (For body centred cubic unit cell)

Avogadro number (N0) =?
straight rho space equals space fraction numerator straight Z space straight x space straight M over denominator straight a cubed space straight x space straight N subscript 0 end fraction

7.875 space straight g space cm to the power of negative 3 end exponent space equals fraction numerator 2 space straight x space left parenthesis 55.845 straight g space mol to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 286.65 space straight x space 10 to the power of negative 10 end exponent space cm right parenthesis space space straight x space left parenthesis 7.874 cm to the power of negative 3 end exponent right parenthesis space straight x space straight N subscript 0 end fraction

straight N subscript 0 space equals space fraction numerator 2 space straight x space left parenthesis 55.845 space straight g space mol to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 286.65 space straight x space 10 to the power of negative 10 end exponent cm right parenthesis cubed space straight x space left parenthesis 7.874 space cm to the power of negative 3 end exponent right parenthesis end fraction space equals space 6.022 space straight x space 10 to the power of 23 space mol to the power of negative 1 end exponent