Complete Question
The complete question is shown on the first uploaded image Â
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]
=> Â Â Â Â Â Â Â Mass of Tungsten (M) = Â [tex]= \frac{184 * 4 }{6.023*10^{23}}[/tex]
=> Â Â Â Â Â Â Â Mass of Tungsten (M) = Â [tex]= 1.222*10^{-21} \ g[/tex]
 Now Â
   The density of  Tungsten is Â
         [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]
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