Question

What is the mass of one mole of electrons?

Answer 1

Hint: One mole is equal to $\text{6}\text{.02214179 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}$ atoms, or other elementary units such as molecules.

Step by step solution:
$\text{1 electron weighs approximately 9}\text{.11 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-31}}}\text{Kg}$
$\text{1 mole weighs X Kg}$
$\text{1 mole is 6}\text{.022}\times \text{1}{{\text{0}}^{\text{23}}}\text{particles}$
So, we put $\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}$in the equation instead.
$\text{X = 9}\text{.11 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-31}}}\text{ }\times \text{ 1 }\!\!\times\!\!\text{ 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{= 5}\text{.48 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}\text{ Kg}$
Mass of $\text{6}\text{.02214076 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}$ electrons $\text{= 548}\text{.1 }\!\!\mu\!\!\text{ g}$

Additional Information:
We should know that mole in chemistry is the standard scientific unit which can be used for measuring large quantities of very small entities for example atoms, molecules or other specified particles.
The Avogadro constant is the proportionality factor that relates the number of constituent particles in a sample with the amount of substance in that sample.
The mass of the electron is $\dfrac{1}{1836}$th the mass of a proton.
The number of moles in a given sample of an element or compound can be calculated by dividing the total mass of the sample by the molar mass of the element or compound.
The mass of 1 mole of an element will be equal to its atomic mass in grams.
For all: $\text{Mole = }\dfrac{\text{Mass}}{\text{Molar Mass}}\text{ }$
For gases: $\text{Mole = }\dfrac{\text{Volume}}{\text{Molar Volume}}\text{ }$ (here molar volume is $\text{24}\text{.0 d}{{\text{m}}^{\text{3}}}$ at r.t.p and $\text{22}\text{.4 d}{{\text{m}}^{\text{3}}}$ at S.T.P)

For solutions:
$\text{Concentration }\left( \text{g / d}{{\text{m}}^{\text{3}}} \right)\text{ = }\dfrac{\text{mass }\left( \text{g} \right)}{\text{volume }\left( \text{d}{{\text{m}}^{\text{3}}} \right)}\text{ }$
$\text{Concentration }\left( \text{mole / d}{{\text{m}}^{\text{3}}} \right)\text{ = }\dfrac{\text{mole}}{\text{volume }\left( \text{d}{{\text{m}}^{\text{3}}} \right)}\text{ }$
$\text{Concentration }\left( \text{g / d}{{\text{m}}^{\text{3}}} \right)\text{ = Concentration }\left( \text{mole / d}{{\text{m}}^{\text{3}}} \right)\text{ }\!\!\times\!\!\text{ molar mass}$

Note: The number $\text{6}\text{.02214076 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}$ was chosen so that the mass of one mole of a chemical compound in grams is numerically equal to the average mass of one molecule of the compound, in Daltons. This number is also known as Avogadro number.