Question

The face centred cubic cell of platinum has a length of $0.392nm$. Calculate the density of platinum $\left(g/c{m}^3\right):$

(Atomic weight : Pt=195)

• A. 20.9
• B. 20.4
• C. 19.6
• D. 21

Answer 1

The correct option is D 21

$Density,d=\frac{Z\times M}{N_A\times a^3}$

Here, Z = No. of atoms in fcc crystal = 4

$N_a=$ Avagadro No. $=6\times {10}^{23}mo{l}^{-1}$, $a=0.392nm=0.392\times {10}^{-7}cm$

$M=195g/mol$

$d=\frac{4\times 195}{N_a\times {(0.392\times {10}^{-7})}^3}$

$d=21\frac{g}{c{m}^3}$
Option D is correct.