Question

Estimate the average mass density of a sodium atom assuming its size to be about $2.5 \AA$. (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

Hii dear,

# Explained answer -
Given is
Radius of sodium atom, r = (1/2)×2.5 Å = 1.25×10^-10 m
Volume of sodium atom,
V = (4/3)πr^3
= (4/3)×3.14×(1.25×10^-10)^3
= 8.2×10^-30 m^3

One mole of sodium contains 6.023×10^23 atoms and has a mass of 23 g.
∴ Mass of one atom
m1 = 23×10^-3 / 6.023×10^23 Kg
= 3.8×10^-26

Density of sodium atom,
ρ = m1 / V
Substituting the value from above, we get
ρ = 4.67×10^-3 Kgm^-3.

Order of magnitude for calculated value is 10^-3 and that for known value is 10^3. i.e. Different

This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.

Hope that was useful...

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.