Question

3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm. The density of the metal is 5.30 g/cm3. Assume that 68% of the unit cell is occupied by Ba atoms. The molar mass of barium is 137.3 g/mol. Using this information, calculate Avogadro’s number. Show your calculation procedure that allows you to derive Avogadro’s number. Your answer must show six digits after the decimal point (i.e., 6.pppx1023) that is not necessarily the same as the known value. By showing your calculation-result down to six digits after the decimal point, you showcase that you did calculate the number, instead of simply adopting the known Avogadro’s number available in open resources.

Answer 1

Answer:

The Avogadro's  number is $N_A = 6.02289 *10^{23}$

Explanation:

From the question we are told that

   The edge length is  $L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}$

    The density of the metal is $\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3$

     The molar mass of  Ba is  $Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol$

     

Generally the volume of a unit cell is  

       $V = L^3$

substituting value

        $V = [5.02 *10^{-10}]^3$

         $V = 1.265*10^{-28}\ m^3$  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

The volume of barium atom is  

        $V_a = \frac{V}{2} * 0.68$

substituting value

        $V_a = \frac{ 1.265*10^{-28}}{2} * 0.68$

        $V_a = 4.301 *10^{-29} \ m^3$

The Molar mass of barium is mathematically represented as

      $Z = N_A V_a * \rho$

Where $N_A$ is the Avogadro's number

 So  

      $N_A = \frac{ Z}{ V_a * \rho}$

substituting value

     $N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}$

     $N_A = 6.02289 *10^{23}$