Solveeit Logo

Question

Question: How many moles of electrons weigh one kilogram? A) \(6.023 \times {10^{23}}moles\) B) \(\dfrac{...

How many moles of electrons weigh one kilogram?
A) 6.023×1023moles6.023 \times {10^{23}}moles
B) 19.108×1031moles\dfrac{1}{{9.108}} \times {10^{31}}moles
C) 6.0239.108×1054moles\dfrac{{6.023}}{{9.108}} \times {10^{54}}moles
D) 19.108×6.023×108moles\dfrac{1}{{9.108 \times 6.023}} \times {10^8}moles

Explanation

Solution

To calculate number of moles of electrons in one kilogram, you should know these values.
Mass of an electron = 9.108×1031kg9.108 \times {10^{ - 31}} kg and, number of electrons in 1 mole = 6.023×1023electrons6.023 \times {10^{23}} electrons.
First, calculate the mass of one mole of electrons and then calculate the number of moles of electrons present in one kilogram by unitary method.

Complete step by step answer:
There are 6.023×1023electrons6.023 \times {10^{23}} electrons (Avogadro number,NA{N_A} ) present in one mole of a substance and mass of one electron is 9.108×1031kg9.108 \times {10^{ - 31}}kg.
Therefore, mass of one mole of electrons = mass of one electron ×\times number of electrons in one mole
\Rightarrow mass of one mole of electrons = (9.108×1031)×(6.022×1023)(9.108 \times {10^{ - 31}}) \times (6.022 \times {10^{23}}) kg
We are asked to find the number of electrons present in one kilogram.
Therefore, total mass given = 1 kilogram
(9.108×1031)×(6.022×1023)(9.108 \times {10^{ - 31}}) \times (6.022 \times {10^{23}}) kg contains = 1 mole of electrons
1 kg will contain = 1(9.108×1031)×(6.022×1023)\dfrac{1}{{(9.108 \times {{10}^{ - 31}}) \times (6.022 \times {{10}^{23}})}} moles of electrons.
Or, 1 kg will contain = 19.108×6.022×108\dfrac{1}{{9.108 \times 6.022}} \times {10^8} moles of electrons.
Thus, 19.108×6.022×108\dfrac{1}{{9.108 \times 6.022}} \times {10^8} moles of electrons will weigh one kilogram.
So, the correct answer is “Option D”.

Additional Information:
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in exactly 12 g (or 0.012 kg) of the carbon-12 isotope. Number of entities or electrons/atoms/molecules/ions present in one mole is equal to the Avogadro constant or Avogadro number (NA{N_A}) which is 6.023×1023particles/mol6.023 \times {10^{23}}particles/mol.

Note: Always remember values of these physical constants:
Charge of an electron= 1.602×1019coulomb1.602 \times {10^{ - 19}}coulomb ,
Mass of an electron = 9.108×1031kg9.108 \times {10^{ - 31}}kg ,
Avogadro constant (NA{N_A}) = 6.023×1023particles/mol6.023 \times {10^{23}}particles/mol,
Charge to mass ratio of an electron (e/me/m )= 1.76×1011coulomb/kg1.76 \times {10^{11}}coulomb/kg .