Question

An imaginary element with BCC structure and has an atomic radius of 0.17 nm, with a molar mass of 56.08 g/mol. What is the density of this element in g/cc? hint: you will need Avogadro's number and you will need to convert the given radius to cm.

Answer 1

Answer: The density of the given element is
`3.07g/cm^3`

Step-by-step explanation:

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

`R=(√(3)a)/(4)`

where,

R = radius of the lattice = 0.17 nm

a = edge length = ?

Putting values in above equation, we get:

`0.17=(√(3)* a)/(4)\\\\a=(0.17* 4)/(√(3))=0.393nm`

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

`\rho=(Z* M)/(N_(A)* a^(3))`

where,

`\rho` = density

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

M = atomic mass of metal = 56.08 g/mol

`N_(A)` = Avogadro's number =
`6.022* 10^(23)`

a = edge length of unit cell =
`0.393nm=3.93* 10^(-8)cm` (Conversion factor:
`1cm=10^(7)nm` )

Putting values in above equation, we get:

`\rho=(2* 56.08)/(6.022* 10^(23)* (3.93* 10^(-8))^3)\\\\\rho=3.07g/cm^3`

Hence, the density of the given element is
`3.07g/cm^3`