Question

Question 5 Tungsten is a solid phase of tungsten still unknown to science. The only difference between it and ordinary tungsten is that Tungsten forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Tungsten .

Answer 1

Complete Question

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Answer:

The density is  $\rho = 21.1 \ g/cm^3$

Explanation:

From the question we are told that

      The lattice constant is  $a = 0.387 nm = 0.387 *10^{-9} \ m$

Generally the volume of the unit cell is  $V = a^3$

                                                             =>   $V = [0.387 *10^{-9}]^2$

                                                                   $V = 5.796 *10^{-29} \ m^3$

Converting to  $cm^3$   We have  $5.796 *10^{-29} * 1000000 = 5.796 *10^{-23} cm^3$

The molar mass of Tungsten is constant with a value  $Z = 184 g/mol$

One mole of Tungsten contains  $6.022*10^{23}$ unit cells

    Where $6.022*10^{23}$  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) =  $= \frac{Z * n }{1 \ mole \ of \ Tungsten}$

=>              Mass of Tungsten (M) =  $= \frac{184 * 4 }{6.023*10^{23}}$

=>              Mass of Tungsten (M) =  $= 1.222*10^{-21} \ g$

 Now  

      The density of  Tungsten is  

                  $\rho = \frac{M}{V}$

substituting values

                $\rho = \frac{1.222*10^{-21}}{5.796*10^{-23}}$

                $\rho = 21.1 \ g/cm^3$