Question

Metallic gold crystallizes in a fcc lattice and has a density of 19.3 g cm -3 . Calculate the radius of
gold atom. (Atomic mass of gold=197 u and Na = 6.022 X 10 -23 mol -1 )
chemistry class 12

Answer 1

Answer:

Correct option is

A

19.4g/cm

3

, 143.9 pm

For an FCC crystal, atoms are in contact along the diagonal of the unit cell.

4r=

2

a, where a is the edge length of unit cell and r is the radius of atom.

Substituting values in the above expression, we get

r=

4

2

×407

=143.9 pm

Number of atoms per unit cell (FCC) =4

Mass of 1 unit cell =197×4=788amu=1.66×10

−27

×788kg=1.30808×10

−24

kg

Volume of 1 unit cell =(407×10

−12

)

3

m

3

=6.741×10

−29

m

3

=2.824×10

−28

m

3

Density=

6.741×10

−29

1.30808×10

−24

=19404.83kg/m

3

=19.4g/cm

3

Explanation:

hope it is helpful pls make me Brainlist