Question

An element forms a body-centered cubic crystalline substance. The edge length of the unit cell is 287 pm and the density of the crystal is 7.92 g/cm3. Calculate the atomic weight of the substance.

Answer 1

Answer:

The atomic weight of the substance is 2.4837 X 10⁻³ amu

Explanation:

For a body-centered cubic structure, the edge length is given as

$X = \frac{4}{\sqrt{3}} *R$

Given edge length  in this problem is 287 pm

$R = \frac{X*\sqrt{3}}{4} = \frac{287 pm*\sqrt{3}}{4} = 124.27 pm$

Volume of a sphere = $\frac{4}{3}\pi R^3$ = 1.333π*(124.27 X 10⁻¹²)³ = 5.206 X 10⁻³⁴ m³ X 10⁶ cm³ = 5.206 X 10⁻²⁸ cm³

Mass = density X volume

Mass = (7.92 g/cm³) *(5.206 X 10⁻²⁸ cm³) = 41.23 10⁻²⁸ g

1.66 x 10⁻²⁴g =  1 amu

41.23 10⁻²⁸ g = 2.4837 X 10⁻³ amu

Therefore, the atomic weight of the substance is 2.4837 X 10⁻³ amu