Question

Given that a certain Aluminium metal consists of face-centered cubic unit cells and has a density is 2.70 g/cm³, calculate (and answer with) the number* of picometers in the approximate radius of these Aluminium atoms

Answer 1

Answer 2

Calculation  :-

 Given:

• Density (ρ) of Aluminium = 2.70 g/cm³

• FCC unit cell structure (4 atoms per unit cell).

• 
  

1. Steps:

   a) Molar Mass & Volume:

   b) Edge Length (a):

   c) Radius (r) in FCC:

• In FCC, atoms touch along the face diagonal:

  $$4r=a\sqrt{2}⟹r=\frac{a\sqrt{2}}{4}$$

• Substitute $a=405\text{ pm}$:

  $$r=\frac{405\times 1.414}{4}=143\text{ pm}$$

• Rearrange density formula to solve for volume (V):

  $$V=\frac{\text{Mass}}{\rho }=\frac{1.792\times 10^{-22}\text{ g}}{2.70{\text{ g/cm}}^3}=6.637\times 10^{-23}{\text{ cm}}^3$$

• Take cube root to find edge length (a):

  $$a=\sqrt[3]{6.637\times 10^{-23}{\text{ cm}}^3}=4.05\times 10^{-8}\text{ cm}=405\text{ pm}$$

• Molar mass of Al (M) = 27 g/mol.

• Use density formula:

  $$\rho =\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$$

• Mass of unit cell = 4 atoms × (27 g/mol) / (6.022 × 10²³ atoms/mol)                                         = 1.792 × 10⁻²² g.

• 
  

• Answer is :
  The radius of an Aluminium atom is 143 pm .