Estimate the average atomic mass density of a sodium atom, assuming its size to be 2.5 Å. Compare it with 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

It is given that radius of sodium atom, `R=2.5 A=2.5xx10^(-10) m`
Volume of one mole atom of sodium, `V=NA. 4//3 pi R^(3)`
`V=6.023xx10^(23)xx-4//3xx3.14xx(2.5xx10^(-10))^(3) m^(3)` and mass of one mole atom of sodium, `M=23 g=23xx10^(-3) kg`
`:.` Average mass density of sodium atom, `p=M//V`
`=(23xx10^(-3)//6.023xx10^(23)xx4//3xx3.14xx(2.5xx10^(-10)))`
`=6.96xx10^(2) kg m^(-3)=0.7xx10^(-3) kg m^(-3)`
the density of sodium in its crystalline phase `=970 kg m^(-3)`
`=0.97xx10^(3) kg m^(-3)`
Obviously the two densities are of the same order of magnitude `(=10^(3) kg m^(-3))`. It is on account of the fact that in solid phase atoms are tightly packed and so the atomic mass density is close to the mass density of solid.