The cubic unit cell of aluminium has an edge length of 405 pm. If its density is 2.7 gcm−3, what is the type of unit cell? (Note: Molar mass = 27 g/mol)
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Face centered cube
Explanation
The density is given by the formula z*M/a^3*NA
Given a = 405ppm, Density ρ = 27gcm^−3 and Molar mass = 27 g/mol, we need to determine z.
Avagadro's number Na =6.022 × 10^23 /mol
⇒z = ρa^3Na/M =[ 27×(405×10^−10)^3 ×6.022×10^23] / 27 = 4
⇒z = 4→ that the unit cell is a face-centered cube.
Explanation
The density is given by the formula z*M/a^3*NA
Given a = 405ppm, Density ρ = 27gcm^−3 and Molar mass = 27 g/mol, we need to determine z.
Avagadro's number Na =6.022 × 10^23 /mol
⇒z = ρa^3Na/M =[ 27×(405×10^−10)^3 ×6.022×10^23] / 27 = 4
⇒z = 4→ that the unit cell is a face-centered cube.
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