Question

Estimate the average mass density of sodium atom assuming it's size to be about 2.5 angstrom( Use the known values of Avogadro's number & atomic mass of sodium) Compare it with density of sodium in its crystalline phase: 970 kg / m³.Are the two densities of the same order of magnitude. If, so why?

Answer 1

Diameter of sodium atom = Size of sodium atom = 2.5 Å
Radius of sodium atom, r = (1/2) × 2.5 Å = 1.25 Å = 1.25 × 10-10 m
Volume of sodium atom, V = (4/3) π r3
= (4/3) × 3.14 × (1.25 × 10-10)3 = VSodium
According to the Avogadro hypothesis, one mole of sodium contains 6.023 × 1023 atoms and has a mass of 23 g or 23 × 10–3 kg.
∴ Mass of one atom = 23 × 10-3 / 6.023 × 1023 Kg = m1
Density of sodium atom, ρ = m1 / VSodium
Substituting the value from above, we get
Density of sodium atom, ρ =4.67 × 10-3 Kg m-3
It is given that the density of sodium in crystalline phase is 970 kg m–3.
Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in 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.