#### Question

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 density of sodium in its crystalline phase: 970 kg m^{–3}. Are the two densities of the same order of magnitude? If so, why?

#### Solution 1

Diameter of sodium atom = Size of sodium atom = 2.5 Å

Radius of sodium atom, r = 1/2 xx 2.5 Å = 1.25 Å

= 1.25 × 10^{–10} m

Volume of sodium atom, `*V* = 4/3 pir^3`

=`4/3 xx 3.14 xx(1.25 xx 10^(-10))^3`

According to the Avogadro hypothesis, one mole of sodium contains 6.023 × 10^{23} atoms and has a mass of 23 g or 23 × 10^{–3} kg.

∴ Mass of one atom = `(23xx10^(-3))/(6.023 xx 10^(23)) kg`

Density of sodiuym atom,p = `((23xx10^(-3))/(6.023xx10^(23)))/(4/3xx3.14xx(1.25xx10^(-10))^3)` = `4.67 xx 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.

#### Solution 2

It is given that radius of sodium atom, R = 2.5 A = 2.5 x 10^{-10} m

Volume of one mole atom of sodium, V = NA .4/3 π R^{3}

V = 6.023 x 10^{23} x –4/3 x 3.14 x (2.5 x 10^{-10})^{3} m^{3}and mass of one mole atom of sodium, M = 23 g = 23 x 10^{-3} kg

∴ Average mass density of sodium atom, p = M/V

=(23 x 10^{-3}/6.023 x 10^{23} x 4/3 x 3.14 x (2.5 x 10^{-10}))

= 6.96 x 10^{2} kg m^{-3} = 0.7 x 10^{-3} kg m^{-3}

The density of sodium in its crystalline phase = 970 kg m^{-3}

= 0.97 x 10^{3} kg m^{-3}

Obviously the two densities are of the same order of magnitude (= 10^{3} kg m^{-3}). It is on account of the fact that in solid phase atoms are tightly packed and so the atomic mass density is close to the mass density of solid.