Question

Calculate the mass and charge of one mole of electrons.

Answer 1

One mole of electrons contains an Avogadro number of electrons. Use the standard values of mass and charge of one electron and calculate the mass and charge of one mole of electrons.

Complete Step by step answer: An atom consists of fundamental particles such as electron, proton and neutron. An electron is a negatively charged subatomic particle. The electron was discovered by J.J. Thomson from the experiments carried out on cathode rays.
As we know that an electron carries a negligible mass of $9.1 \times {10^{ - 31}}{\text{kg}}$. Its mass is 1/1837 times the mass of a hydrogen atom.
An electron carries a charge of${\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C}}$.
So, we know the mass and charge of an electron.
To calculate the mass and charge of one mole of electrons we have to determine the number of electrons present in one mole of electrons as follows:
1 mole of any substance always contains an Avogadro number of particles.
Avogadro number = ${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}$
So, we can say that
${\text{1 mole of electrons = 6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{electrons}}$
Now, we know the number of electrons present in 1 mole of electrons and the mass and charge of an electron.
So, we can calculate the mass of one mole of electrons as follows:
Mass of an electron = $9.1 \times {10^{ - 31}}{\text{kg}}$
So, ${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ electrons }} \times \dfrac{{9.1 \times {{10}^{ - 31}}{\text{kg}}}}{{1{\text{ electron}}}} = {\text{ 5}}{\text{.48}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}{\text{kg}}$
Thus, the mass of one mole of electrons is ${\text{5}}{\text{.48}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}{\text{kg}}$.
Now, using the charge of an electron we can calculate the charge of one mole of an electron as follows:
Charge of an electron = ${\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C}}$
So, ${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ electrons }} \times \dfrac{{{\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{{\text{ - 19}}}}{\text{C}}}}{{1{\text{ electron}}}} = {\text{ 9}}{\text{.64}} \times {\text{1}}{{\text{0}}^{\text{4}}}{\text{C}}$
Thus, the charge of one mole of electrons is${\text{ 9}}{\text{.64}} \times {\text{1}}{{\text{0}}^{\text{4}}}{\text{C}}$.
Hence, the mass and charge of one mole of electrons is ${\text{5}}{\text{.48}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}{\text{kg}}$ and ${\text{ 9}}{\text{.64}} \times {\text{1}}{{\text{0}}^{\text{4}}}{\text{C}}$ respectively.

Note: Out of three subatomic particles the mass of an electron is very negligible. Avogadro number indicates the number of particles present in 1 mole of a substance. It has a constant value of ${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}$.