Answer:
92.4%
Explanation:
The volume per cubic meter of sodium preoccupy by the sodium ions is
[tex]V_{Na} = n \times V[/tex]
where;
volume (V) of each Na atom = [tex]\dfrac{ 4}{3} \pi r^3[/tex]
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
∴
[tex]V = \bigg [ \dfrac{4}{3} \pi r^3 \bigg ] \bigg [\dfrac{MN_A}{m} \bigg ][/tex]
[tex]V = \bigg [ \dfrac{4}{3} \pi (97.3 \times 10^{-12} )^3 \bigg ] \bigg [\dfrac{ (911 kg/m^3 (m^3))(6.023\times 10^{23})}{27.7 \times10^{-3}\ kg/mol } \bigg ][/tex]
V = 0.07643 m³
The fraction of available conduction electrons are;
= (1 - V)
= 1 - 0.07643
= 0.92357
≅ 92.4%
