The unit cell of aluminium is a cube with an edge length of $405 p m$ . The density of aluminium is $2.70 g c m^{- 3}$ . What type of unit cell of aluminium is?

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Solution

$ρ = \frac{Z \times M_{A}}{N_{A} \times a^{3}}$
$Z = ?$ , $ρ = 2.7 g / c m^{3}$ , $M_{A} = 27$ , $a = 4.05 ˚ A$
$2.7 = \frac{Z \times 27}{6.023 \times 10^{23} \times 66.43 \times 10^{- 24}}$

$\Rightarrow Z = 4$

Therefore it is FCC unit cell.

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An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m⁻³, what is the nature of the cubic unit cell?

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The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is 2.70 g cm−3. What type of unit cell of aluminium is?

ρ=Z×MANA×a3Z=? , ρ=2.7g/cm3 , MA=27 , a=4.05˚A2.7=Z×276.023×1023×66.43×10−24⇒Z=4Therefore it is FCC unit cell.

The unit cell of aluminium is a cube with an edge length of $405 p m$ . The density of aluminium is $2.70 g c m^{- 3}$ . What type of unit cell of aluminium is?

$ρ = \frac{Z \times M_{A}}{N_{A} \times a^{3}}$
$Z = ?$ , $ρ = 2.7 g / c m^{3}$ , $M_{A} = 27$ , $a = 4.05 ˚ A$
$2.7 = \frac{Z \times 27}{6.023 \times 10^{23} \times 66.43 \times 10^{- 24}}$

$\Rightarrow Z = 4$

Therefore it is FCC unit cell.