Iron has a body centred cubic 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 μ )
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Hey !!
We need to apply the density formula
d = Z × M / a³ × Na
Now, bcc lattice Z = 2
7.874 = 2 × 55.845 g mol⁻¹ / (286.65 × 10⁻¹⁰ cm³ ) Na
= Na = 2 × 55.845 g mol⁻¹ / (286.65 × 10⁻¹⁰ cm)³ ×7.874 g cm⁻³
= Na = 6.02 × 10²³ mol⁻¹
Good luck !!
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