Question

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

Answer 1

Given:

The diameter of the sodium atom is $=2.5\stackrel{o}{A}$

The volume of the sodium atom can be obtained as:

$V=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times 3.14\times (1.25\times {10}^{-10}{)}^3$

$1mole$ of sodium contains $6.023\times {10}^{23}$ atoms.

The mass of $1mole$ of sodium is $23g$ or $23\times {10}^{–3}kg$.

The mass of one sodium atom is:

$m_1=\frac{23\times {10}^{-3}}{6.023\times {10}^{23}}Kg$

We know that the density of the atom can be obtained as:

$\rho =\frac{m_1}{V}$

The density of the sodium atom is:

$\rho =\frac{\frac{23\times {10}^{-3}}{6.023\times {10}^{23}}}{\frac{4}{3}\times 3.14\times (1.25\times {10}^{-10}{)}^3}$

$\rho =4.67\times {10}^{-3}Kg{m}^{-3}$

It is given that the density of sodium in the crystalline phase is 970 kg m–3.

Therefore, the densities of sodium in two phases is not of the same order. In the solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.