Question

The face centered cubic cell of platinum has a length of o.392nm. Calculate the density of platinum \[(g/c{m^{3)}}\].
tomic weight of Pt=$195$
(A) 20.9
(B) 20.4
(C) 19.6
(D) 21

Answer 1

Hint: Platinum adopts a face centered cubic crystal structure. This is also known as cubic close packed structure. It consists of 4 equivalent metal atoms in a cubic unit cell.
Formula used:
$d = \dfrac{{Z \times M}}{{{N_a} \times {a^3}}}$
Where, Z = Number of atoms in fcc crystal = 4
M= 195 (given in question)
${N_a}$= Avogadro’s number =$6 \times {10^{23}}mo{l^{ - 1}}$
d=density
a=$0.392nm = 0.39 \times {10^{ - 7}}cm$

Complete step by step solution:
The density of a unit cell is given as the ratio of mass and volume of unit cells.The mass of a unit cell is equal to the product of number of atoms in a unit cell and the mass of each atom in the unit cell.
Mass of unit cell=number of atoms in unit cell$ \times $mass of each atom
$ = z \times m$ ( where z=number of atoms in unit cell and m=mass of each atom)
Mass of atom can be given with the help of Avogadro’s number and molar mass, this is given as:
$\dfrac{M}{{{N_A}}}$
Now, volume of unit cell, $V = {a^3}$
Therefore, the density of unit cell will be $\dfrac{m}{v} = \dfrac{{z \times m}}{{{a^3}}}$
$ = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}$
According to the above given formula,
We will find out the density of platinum.
Now,
$d = \dfrac{{4 \times 195}}{{6 \times {{10}^{23}} \times {{(0.392 \times {{10}^{ - 7}})}^3}}}$
$d = 21gc{m^{ - 3}}$
Therefore,
The density is $21gc{m^{ - 3}}$

Hence, option D is correct.

Note: Platinum is a silvery white metal. It is extremely resistant to tarnishing and corrosion (which makes it a noble metal). It is used in chemical industries as a catalyst for the production of nitric acid, silicon and benzene. It is one of the densest precious metals, followed by gold, mercury, lead and silver.