Question

Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (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 kgm-3. Are the two densities of the same
order of magnitude ? If so, why?​

Answer 1

Answer:

sodium is Na.............

Answer 2

Given:

• The radius of the sodium atom (r) = $2.5 \AA$

Converting it into m we get,

[ $1\AA = 10^{-10}m$ ]

• The volume of the sodium atom = $\frac{4}{3}\pi r^{3}$

Volume  $=\frac{4}{3}\times 3.14 \times (2.5\times 10^{-10})^{3}$

Volume = $65.42 \times 10^{-10}\ m^{3}$

Now,

The numbers of atom in one mole of sodium = Avogadro's number (N)

So,

N = 6.023 × 10²³

Therefore,

The atomic volume of sodium = Volume of one atom of sodium × Number of atoms

The atomic volume of sodium = $65.42 \times 10^{-10} \times 6.023 \times 10^{23}$

The atomic volume of sodium = $3.94 \times 10^{-5}\ m^{3}$

Mass of one mole of sodium = 23g = 23 × 10⁻³ kg

$\boxed{\bf{Average\ mass\ density\ of\ soidum =\frac{Mass}{Volume}}}$

$\rho= \boxed{\frac{23\times10^{-3}}{3.94\times 10^{-5}}} \approx 5.84 \times 10^{2}\ kgm^{-3} \approx \bf 584\ kgm^{-3}$

And,

The density of sodium in the crystalline phase

$= 970 kgm^{-1} = 9.7 \times 10^{2}kgm^{-3]$

So,

The two densities are of the same order of magnitude because, in solid-state, atoms are trightly packed.