Question

A metallic solid with atoms in a face-centered cubic unit cell with an edge length of 392 pm has a density of $\mathbf{21}.\mathbf{45}\mathbf{g}/{\mathbf{cm}}^{\mathbf{3}}$. Calculate the atomic mass and the atomic radius of the metal. Identify the metal.

Short answer

Atomic mass = 194

Atomic radius = 1.3910^-8 cm

Metal is Platinum.

Step-by-step solution

Finding atomic radius in a face-centered cubic unit

$$\begin{gathered} r = atomic radius \\ a = edge length \\ In FCC \\ \sqrt{2}a = 4r \\ \sqrt{2}(3.92\times {10}^{-8}) = 4r \\ r= 1.39\times {10}^{-8} cm \end{gathered}$$

Determining the molar mass

Let the atomic mass be M

$$\begin{gathered} Density = \frac{4 atoms \times \frac{1 mol }{6.022 \times {10}^{23} atoms}\times \frac{M}{mol }}{{(3.92 \times {10}^{-8} cm)}^3} \\ 21.1g/c{m}^3=\frac{4 atoms \times \frac{1 mol }{6.022 \times {10}^{23} atoms}\times \frac{M}{mol }}{{(3.92 \times {10}^{-8} cm)}^3} \\ On putting density value here \\ M = 194 \end{gathered}$$

Determining the atomic radius

The metal is Platinum.