# Q 2.27: Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’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?

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Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: Are the two densities of the same order of magnitude? If so, why? Given, Diameter of sodium = 2.5 $\AA$

Radius, r = $0.5 \times 2.5 � = 1.25 \times 10^{-10}\ m$

Now, Volume occupied by each atom  $V = (4/3)\pi r^3 = (4/3)\times (22/7)\times(1.25\times 10^{-10})^3$

$= 8.18 \times 10^{-30} m^3$

We know, One mole of sodium has $6.023 \times 10^{23}$ atoms and has a mass of  $23 \times 10^{-3} kg$

$\therefore$ Mass of each Sodium atom $= \frac{23\times10^{-3}}{6.023\times10^{23}} kg$

$\therefore$ Density  = Mass/ Volume $= (\frac{23\times10^{-3}}{6.023\times10^{23}}\ kg )/ (8.18 \times 10^{-30} m^3) = 466.8 \times 10^{1}\ kgm^{-3}$

But, the mass density of sodium in its crystalline phase =  $970 \ kg\ m^{-3}$

$\therefore$ The densities are almost of the same order. In the solid phase, atoms are tightly packed and thus interatomic space is very small.

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