Complete Question
The complete question is shown on the first uploaded image
Answer:
The density is [tex]\rho = 21.1 \ g/cm^3[/tex]
Explanation:
From the question we are told that
The lattice constant is [tex]a = 0.387 nm = 0.387 *10^{-9} \ m[/tex]
Generally the volume of the unit cell is [tex]V = a^3[/tex]
=> [tex]V = [0.387 *10^{-9}]^2[/tex]
[tex]V = 5.796 *10^{-29} \ m^3[/tex]
Converting to [tex]cm^3[/tex] We have [tex]5.796 *10^{-29} * 1000000 = 5.796 *10^{-23} cm^3[/tex]
The molar mass of Tungsten is constant with a value [tex]Z = 184 g/mol[/tex]
One mole of Tungsten contains [tex]6.022*10^{23}[/tex] unit cells
Where [tex]6.022*10^{23}[/tex] is a constant for the number of atom in one mole of a substance(Tungsten) which is known as Avogadro's constant
Now for FCC distance the number of atom per unit cell is n = 4
Mass of Tungsten (M) = [tex]= \frac{Z * n }{1 \ mole \ of \ Tungsten}[/tex]
=> Mass of Tungsten (M) = [tex]= \frac{184 * 4 }{6.023*10^{23}}[/tex]
=> Mass of Tungsten (M) = [tex]= 1.222*10^{-21} \ g[/tex]
Now
The density of Tungsten is
[tex]\rho = \frac{M}{V}[/tex]
substituting values
[tex]\rho = \frac{1.222*10^{-21}}{5.796*10^{-23}}[/tex]
[tex]\rho = 21.1 \ g/cm^3[/tex]
