
Calculate the mass and charge of one mole of electrons.
Answer
472.5k+ views
Hint: 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 \[{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\].
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 \[{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\].
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Why was the Vernacular Press Act passed by British class 11 social science CBSE

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

Name the nuclear plant located in Uttar Pradesh class 11 social science CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

How did silk routes link the world Explain with three class 11 social science CBSE

What are the various challenges faced by political class 11 social science CBSE
