QJEEAdv02Chd6a4: JEE Advanced 2002 Chemistry - Class 11 JEEAdv02Chd6a4 Solution: How many moles of electron weigh one kilogram?

To find out how many moles of electrons weigh one kilogram, we need to understand the concept of the mole and the mass of an electron.

A mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, electrons, etc.) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number, $N_A = 6.023 \times 10^{23}$ entities per mole.

The mass of an electron is approximately $9.109 \times 10^{-31}$ kg per electron.

The number of moles ( $n$ ) is given by the equation:

$n = \frac{m}{M}$

where $m$ is the mass of the substance and $M$ is the molar mass of the substance.

In the case of electrons, the mass ( $m$ ) we're considering is 1 kilogram, and we need to find the equivalent mass in terms of moles of electrons.

However, instead of using the molar mass as we would for atoms or molecules, we directly relate the mass of the electrons to Avogadro's number since we're dealing with elementary particles.

The calculation involves finding out how many electrons make up 1 kg, and then calculating how many moles of electrons that represents.

First, find the number of electrons in 1 kg:

$\text{Number of electrons} = \frac{1 \, \text{kg}}{(9.109 \times 10^{-31} \, \text{kg/electron})} = \frac{1}{9.109 \times 10^{-31}}$

Then, to find out how many moles this number of electrons represents, we divide by Avogadro's number:

$\text{Moles of electrons} = \frac{1}{9.109 \times 10^{-31}} \times \frac{1}{6.023 \times 10^{23}}$

Simplifying this gives us:

$\text{Moles of electrons} = \frac{1}{9.109} \times 10^{31} \times \frac{1}{6.023 \times 10^{23}} = \frac{1}{9.109 \times 6.023} \times 10^{8}$

Hence, the option that correctly represents the number of moles of electrons that weigh one kilogram is:

Option D: ${1 \over {9.108 \times 6.023}} \times {10^8}$