Question

Copper, a monovalent metal, has molar mass $63.54g/mol$ and density $8.96g/c{m^3}$ . What is the number density n of conduction electrons in copper.

Answer 1

Hint: The number density is usually used to describe the degree of concentration in space. In chemistry it is usually used for the same quantity or when comparing with other concentrations. It is denoted by the symbol n or ${\rho _N}$.In order to solve this problem first of all we’ll find the number density of copper and than molar mass and after substituting that in $n = \dfrac{{d \times {N_A}}}{m}$ ,we’ll have our answer.

Complete answer:
The metal given to us is copper. It is given that Copper is a monovalent metal hence the no. of electrons in conduction will be one only. To find the number density ‘n’ of copper we will use the formula: $n = \dfrac{d}{M}$
Where d is the density and M is the molar mass of one atom of copper i.e. the mass of one mole of copper. But we are given the Molar Mass of copper, to find the mass of one atom we will divide the Molar mass by Avogadro’s number, since Avogadro’s number is equal to the no. of atoms/molecules in one mole of any compound.
 $M = \dfrac{m}{{{N_A}}}$ (Where m is the molar mass)
Substituting this in the above formula we get,
$n = \dfrac{{d \times {N_A}}}{m}$
The values given to us are: Molar Mass = $63.54g/mol$ , Density = $8.96g/c{m^3}$. Finding the value of n:
$n = \dfrac{{8.96 \times 6.022 \times {{10}^{23}}}}{{63.54}}$
$n = 8.49 \times {10^{22}}c{m^{ - 3}} = 8.49 \times {10^{28}}{m^{ - 3}}$
Hence, the number density n of conduction electrons in copper is $8.49 \times {10^{28}}/{m^3}$

Note:
Alternate way to do this question is converting the molar mass from g/mol to $kg/mol$ and density from $g/c{m^3}$ to $kg/{m^3}$. Remember that, 1 kg = 1000g and therefore, $1g = {10^{ - 3}}kg$. Also 1m = 100 cm, $1cm = {10^{ - 2}}m$. Which means that $1c{m^3} = {({10^{ - 2}})^3} = {10^{ - 6}}{m^3}$.
Upon conversion the values obtained are:
Density = $d = 8.96g/c{m^3} = 8.96 \times \dfrac{{1kg}}{{{{10}^3}kg}} \times \dfrac{{{{10}^6}c{m^3}}}{{1{m^3}}} = 8.96 \times {10^3}kg/m$
Molar mass = $m = 63.54g/mol = 0.06354kg/mol$
Substituting the values in the formula above, we get : $n = \dfrac{{(8.96 \times {{10}^3}kg/{m^3})(6.022 \times {{10}^{23}}mo{l^{ - 1}})}}{{0.06354kg/mol}} = 8.49 \times {10^{29}}/{m^3}$