How many moles of electrons weighs one kilogram?
A. $\dfrac{1}{\left( 9.108\times 6.023 \right)}\times {{10}^{8}}$
B. $\dfrac{1}{9.108\times {{10}^{32}}}$
C. $\dfrac{6.023}{9.108\times {{10}^{54}}}$
D. $\dfrac{1}{\left( 9.108\times 6.023 \right)}\times {{10}^{8}}$

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Hint: We know that one mole is equal to $6.022\times {{10}^{23}}$ atoms, or other elementary units such as molecules. This is the basic information we have to keep in mind while solving the numerical question.

Complete answer:
First, we will consider the mass of one electron, then we will find the number of moles of electrons that 1 kg carried by establishing the relation for cross multiplication as:
\[\dfrac{\text{1 electron}}{\text{weight of 1 electron}}=\dfrac{x\text{ electrons}}{1kg}\]
Thus, we can find the number of electrons present in 1 kg. Then divide them by the Avogadro’s number to find the number of moles.
We know that, mass of one electron = $9.108\times {{10}^{-31}}kg$
Now, the number of electrons that are present in $1kg$ are:
\[x\text{ electrons = }\dfrac{1}{9.108\times {{10}^{-31}}kg}\]
Now we will divide this number by the Avogadro’s number (number of atoms present in one mole of any entity) to get the number of moles of electrons in $1kg$of electrons.
  & \text{No}\text{. of moles of electrons = }\dfrac{\dfrac{1}{9.108\times {{10}^{-31}}}}{6.022\times {{10}^{23}}} \\
 & \text{No}\text{. of moles of electrons = }\dfrac{1}{9.108\times 6.022\times {{10}^{-8}}} \\
 & \text{No}\text{. of moles of electrons = }\dfrac{1}{9.108\times 6.022}\times {{10}^{8}} \\

Therefore, the correct option is ‘Option D’.

Note: The number $6.02214076\times {{10}^{23}}$ is the number of units of an element present in 1 mole of the substance. This number is fixed and does not change for any entity, be it atoms, molecules or electrons and charged particles. In gases, the volume occupied by 1 mole of a gas is also fixed as 22.4 liters and does not vary at STP.