Question

Estimate the average mass density of a sodium atom assuming its size to be about
2.5 Angstrom. (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. Are
the two densities of the same order of magnitude ? If so, why?​

Answer 1

Answer:

l

Mass of each Sodium atom

Density = Mass/ Volume

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

>>>>> ρ=4.67×10 −3 Kgm −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.

Answer 2

$\huge\underline{Explanation•}$

$volume \: of \: Na \: atom \: = \: v \: = \: \frac{4}{3} \pi {r}^{3} \\ \: = \frac{4}{3} \times 3.14 \times ({2.5 \times 10 ^{ - 10} })^{3} \\ = 65.42 \times {10}^{ - 30} m ^{3}$

Now,

$volume \: of \: one \: mole \: of \: Na \: atom \: \: v = na \times v \\ = 6.023 \times {10}^{23} \times 65.42 \times {10}^{ - 30} \: {m}^{3} \\ = 3.95 \times {10}^{ - 5} \: {m}^{3}$

Mass of one mole of Na atom = 23 gm

$density \: of \: Na \: atoms \: = \frac{23 \times {10}^{ - 3} }{3.95 \times {10}^{ - 5} } = 584 \: kg \: {m}^{ - 3}$

Here , 584 and 970 are of same order.

Note:- Density of sodium in crystalline phase little differ from density of atoms because in crystalline molecules are arranged in sequence and in atoms closely packed.