Question

Assume that the total volume of a metal sample is the sum of the volume occupied by the metal ions making up the lattice and the (separate) volume occupied by the conduction electrons. The density and molar mass of the first metal are 911 kg/m3 and 27.7 g/mol, respectively; assume the radius of an ion is 97.3 pm. (a) What percent of the volume of a sample of this metal is occupied by its conduction electrons

Answer 1

92.4%

Step-by-step explanation:

The volume per cubic meter of sodium preoccupy by the sodium ions is

`V_(Na) = n * V`

where;

volume (V) of each Na atom =
`( 4)/(3) \pi r^3`

Radius of each ion = 97.3 pm = 97.3 × 10⁻¹²

no.of atoms in the sample n = mass of the sample / ( molar mass / NA)

mass of the sample per cubic metre = 911 kg/m³

∴

`V = \bigg [ (4)/(3) \pi r^3 \bigg ] \bigg [(MN_A)/(m) \bigg ]`

`V = \bigg [ (4)/(3) \pi (97.3 * 10^(-12) )^3 \bigg ] \bigg [( (911 kg/m^3 (m^3))(6.023* 10^(23)))/(27.7 *10^(-3)\ kg/mol ) \bigg ]`

V = 0.07643 m³

The fraction of available conduction electrons are;

= (1 - V)

= 1 - 0.07643

= 0.92357

≅ 92.4%