Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are <em>4 atoms per unit cell.</em>
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = <em>1.792x10⁻²²g</em>
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
<h3>143pm is the radius of an Al atom</h3>
