How many moles of electrons weigh in one kilogram?

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$\frac{1}{9.108 \times 6.023} \times 10^{8}$

$\frac{6.023}{9.108} \times 10^{54}$

$6.023 \times 10^{23}$

$\frac{1}{9.108} \times 10^{31}$

answer is D.

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Detailed Solution

Mass of one electron is $9.108 \times 10^{- 31} kg$ and the number of electrons present in one mole is $6.023 \times 10^{23} electrons .$

Mass of one mole of electrons determined by multiplying the number of electrons in one mole with the mass of one electron.

$\begin{matrix} Mass of one mole of electron & = & 6.023 \times 10^{23} \times 9.108 \times 10^{- 31} \\ = & 6.023 \times 9.108 \times 10^{- 8} kg \end{matrix}$

In the given problem, the total mass of electron is 1 kg and the mass of one mole of electron is $6.023 \times 9.1 \times 10^{- 8} kg .$ Hence, the total number of moles of electron present in one kg is determined as:

$Number of moles of electrons in one kg = \frac{10^{8}}{9.108 \times 6.023}$