Question

How many moles of electron weighs one kilogram?

• A. $6.023\times {10}^{23}$ moles
• B. $\frac{1}{9.108\times 6.023\times {10}^8}$ moles
• C. $\frac{6.023}{9.108\times {10}^{54}}$ moles
• D. $\frac{1}{9.108\times 0.023\times {10}^{31}}$ moles

Answer 1

The correct option is B

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

Explanation for the correct option:

Option (B):
Step 1: Calculation of one mole of the electron:
The number of electrons present in one mole of a substance =$6.023\times {10}^{23}electrons$ (Avogadro number,$N_A$ )

Also the mass of one electron i=$9.108\times {10}^{-31}kg$
Therefore,

The mass of one mole of electrons can be given as= the mass of one electron ×$\times$ number of electrons in one mole
$\Rightarrow$ mass of one mole of electrons =$(9.108\times {10}^{-31})\times (6.022\times {10}^{23})$ kg

Step 2: Calculation of moles of electron weighing in one kilogram:
We are asked to find the number of electrons present in one kilogram.
We have already calculated,

mass of one mole of electrons =$(9.108\times {10}^{-31})\times (6.022\times {10}^{23})$ kg
$\Rightarrow (9.108\times {10}^{-31})\times (6.022\times {10}^{23})$ kg contains = 1 mole of electrons
$\Rightarrow$1 kg will contain = ${\displaystyle \frac{1}{(9.108\times {10}^{-31})\times (6.022\times {10}^{23})}}$ moles of electrons.
Or, 1 kg will contain = ${\displaystyle \frac{1}{9.108\times 6.022}}\times {10}^8$ moles of electrons.
Thus, the number of moles of electrons weighing in one kilogram= ${\displaystyle \frac{1}{9.108\times 6.022}}\times {10}^8$ moles of electrons

Thus, option B is the correct option.