Question

Sodium metal (atomic weight 22.99 g/cm^3) adopts a body-centered cubic structure with a density of 0.97 g/cm^3. (a) Use this information and Avogrado's number (Na=6.022x10^23) to estimate the atomic radius of sodium. (b) If it didn't react so vigorously, sodium could float on water. Use the answer from part (a) to estimate the density of Na if its structure were that of a cubic close-packed metal. Would it still float on the water?

Answer 1

Answer:

The atomic radius of sodium is 185.59 pm.

Sodium metal will float in water.

Explanation:

a) To calculate the density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density  = $0.97 g/cm^3$

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal = 22.99 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = ?

Putting values in above equation, we get:

$0.97 g/cm^3=\frac{2\times 22.99}{6.022\times 10^{23}\times (a)^3}$

$a=4.286\times 10^{-8} cm=4.286\times 10^{-8}\times 10^{10} pm = 428.6 pm$

$1 cm = 10^{10} pm$

To calculate the radius, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 428.6 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 428.6}{4}=185.59 pm$

The atomic radius of sodium is 185.59 pm.

b)

Density of water = D = 1 g/mL = $1 g/cm^3$

( $1 mL = 1cm^3$)

Density of the sodium metal = d =$0.97 g/cm^3$

D > d ( float)

Since, the density of the sodium metal is less than the water's density which means that sodium metal will float in water.