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

The atomic radius of sodium in a BCC structure can be calculated using its density and Avogadro's number. The estimated radius can then assist in inferring the density if sodium adopted a CCP structure. Despite the denser packing in CCP, sodium would likely still float on water due to its low atomic mass.

Step-by-step explanation:

To estimate the atomic radius of sodium in a body-centered cubic (BCC) structure, we utilize its given density and Avogadro's number. Since a BCC structure has 2 atoms per unit cell and sodium's atomic mass is approximately 22.99 g/mol, we can calculate:

The volume of a Na atom in cm³/mol using the formula V = mass / density.

Since there are 2 atoms per unit cell: V(atom) = V(cell) / 2.

The length of the cube's edge, a, using the volume of the cell: a = V(cell)³.

The atomic radius (r) using the relationship a = 4r / √3 in a BCC lattice.

For part (b), we consider the cubic close-packed (CCP) structure, which has 4 atoms per unit cell. Assuming that the atomic radius and mass remain constant, we calculate the change in density. Since the CCP structure is more efficient in packing atoms, the density will be higher. Despite that, the resulting density would still likely be less than that of water (1 g/cm³), meaning sodium would still float.

Answer 2

The atomic radius of sodium is 185.59 pm.

Sodium metal will float in water.

Step-by-step explanation:

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

`\rho=(Z* M)/(N_(A)* 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* 10^(23)`

a = edge length of unit cell = ?

Putting values in above equation, we get:

`0.97 g/cm^3=(2* 22.99)/(6.022* 10^(23)* (a)^3)`

`a=4.286* 10^(-8) cm=4.286* 10^(-8)* 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=(√(3)a)/(4)`

where,

R = radius of the lattice = ?

a = edge length = 428.6 pm

Putting values in above equation, we get:

`R=(√(3)* 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.