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?

Asked by Abhisek | 1 year ago |  195

##### Solution :-

The diameter of sodium= 2.5 A = 2.5 x 10-10 m

Therefore, the radius is 1.25 x 10-10 m

Volume of sodium atom, V= ($$\dfrac{4}{3}$$)πr3

= ($$\dfrac{4}{3}$$) x ($$\dfrac{22}{7}$$) x (1.25 x 10-103= 8.177 x 10-30 m3

Mass of one mole atom of sodium = 23 g = 23 x 10-3 kg

1 mole of sodium contains 6.023 x 1023 atoms

Therefore, the mass of one sodium atom, M

$$\dfrac{ 23 \times 10^{-3}}{6.023 \times 10^{23}}$$= 3.818 x 10-26 kg

Atomic mass density of sodium, ρ= $$\dfrac{M}{V}$$
=$$\dfrac{ 3.818 \times 10^{-26}}{8.177 \times 10^{-30}}$$

= 0.46692 x 104= 4669.2 kg m-3
The density of sodium in its solid state is 4669.2  kg m-3 but in the crystalline phase, density is 970 kg m-3. Hence, both are in a different order. In solid-phase, atoms are tightly packed but in the crystalline phase, atoms arrange a sequence which contains void. So, density in solid-phase is greater than in the crystalline phase.

Answered by Pragya Singh | 1 year ago

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