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Density

Iron has a bo...

Question

Iron has a body-centered cubic unit cell with a cell dimension of $286.65 p m$ . The density of iron is $7.874 g c m^{- 3}$ . Use this information to calculate Avogadro's number. (At. Mass of $F e = 55.845 a m u$ )

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Solution

$\Rightarrow G i v e n b c c u n i t c e l l \to n o . o f a t o m s i n u n i t c e l l = 2 a = 286.65 p m d = 7.87 g {c m}^{- 3} a t o m i c m a s s o f F e = 56.0 μ d = \frac{Z \times M}{a^{3} \times N^{0}} [w h e r e d = d e n s i t y, z = n o . o f a t o m s i n u n i t c e l l, a^{3} = F e v o l u m e, M = a t o m i c m a s s, N_{0} = a v o g a d r o n o .] N^{o} = \frac{Z \times M}{a^{3} \times d} = \frac{2 \times 56}{(286.65 \times 10^{- 10}) \times (7.87)} = 6.022 \times 10^{23} {m o l}^{- 1}$

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