Estimate the average mass density of a sodium atom assuming its size to be about 2.5Å. (Use the known values of Avagardo's number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : `970 kg m^(-3)`. Are the two densities of the same order of magnitude ? If so, why ?

Explanation

Radius of sodium atom `r=2.5Å`
`=2.5xx10^(-10)m`
Volume of sodium atom `V.=(4)/(3)pir^(3)`
`=(4)/(3)xx3.14xx(2.5xx10^(-10))^(3)`
`65.416xx10^(-30)`
`~~65.42xx10^(-30)m^(3)`
No. of atoms in 1 mole `N=6.023xx10^(23)`
volume of 1 mole of sodium
V=NV.
`=65.42xx10^(-30)xx6.023xx10^(23)`
`=394.02xx10^(-7)`
`~~3.94xx10^(-5)m^(3)`
Mass of 1 mole of sodium = sodium mass
`M=23g=23xx10^(-3)kg`
Average density of 1 mole atoms of sodium
`rho=(M)/(V)`
`:.rho=(23xx10^(-3))/(3.94xx10^(-5))`
`=5.8375xx10^(2)`
`:.rho~~584kgm^(-3)`
Density of sodium in crystalline form
`rho~~970kgm^(-3)`
Ratio `(rho)/(rho.)=(584)/(970)`
=0.6020
0.6
Yes, both densities are in same no. of power. Because atoms are bound strongly in solid form. Therefore, atomic density is close to density of solid.