Answer:
The Avogadro's number is [tex]N_A = 6.02289 *10^{23}[/tex]
Explanation:
From the question we are told that
The edge length is [tex]L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}[/tex]
The density of the metal is [tex]\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[/tex]
The molar mass of Ba is [tex]Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol[/tex]
Generally the volume of a unit cell is
[tex]V = L^3[/tex]
substituting value
[tex]V = [5.02 *10^{-10}]^3[/tex]
[tex]V = 1.265*10^{-28}\ m^3[/tex]
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
[tex]V_a = \frac{V}{2} * 0.68[/tex]
substituting value
[tex]V_a = \frac{ 1.265*10^{-28}}{2} * 0.68[/tex]
[tex]V_a = 4.301 *10^{-29} \ m^3[/tex]
The Molar mass of barium is mathematically represented as
[tex]Z = N_A V_a * \rho[/tex]
Where [tex]N_A[/tex] is the Avogadro's number
So
[tex]N_A = \frac{ Z}{ V_a * \rho}[/tex]
substituting value
[tex]N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}[/tex]
[tex]N_A = 6.02289 *10^{23}[/tex]
