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# 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. $\dfrac{1}{\left( 9.108\times 6.023 \right)}\times {{10}^{8}}$

Last updated date: 19th Jun 2024
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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.

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.
\begin{align} & \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}} \\ \end{align}

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.