
How many moles of electrons weigh one kilogram?
(Mass of electron = ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ kg}}$; Avogadro number, = \[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\]
(A) \[\dfrac{1}{{{\text{9}}{\text{.108 x 6}}{\text{.023 }}}}{\text{x1}}{{\text{0}}^{\text{8}}}\]
(B) \[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\]
(C) \[\dfrac{1}{{{\text{9}}{\text{.108}}}}{\text{x 1}}{{\text{0}}^{{\text{31}}}}\]
(D) \[\dfrac{{{\text{6}}{\text{.023}}}}{{{\text{9}}{\text{.108 }}}}{\text{x1}}{{\text{0}}^{54}}\]
Answer
224.7k+ views
Hint: One mole of any substance has Avogadro number of atoms or molecules or ions. Avogadro number,\[{\text{\;}}{{\text{N}}_{\text{A}}}\]= \[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\]
Complete step by step answer: It is given that mass of one electron = ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ kg}}$
We know that one mole of any substance has an Avogadro number of atoms or molecules or ions. Avogadro number,\[{\text{\;}}{{\text{N}}_{\text{A}}}\]= \[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\]
⟹Mass of one mole of electrons = ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ x 6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}$kg
Since in the options we can see that the numbers are not in simplified form so we do not perform the multiplication we can keep it as such till the end.
Now, we need to find the number of moles of electrons that weigh one kilogram(kg)
⟹ The number of mole of electrons that weigh one kilogram = $\dfrac{1}{{{\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ x 6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}}}$
Now let’s just simplify the power terms, by doing so we get,
⟹The number of mole of electrons that weigh one kilogram = $\dfrac{1}{{{\text{9}}{\text{.108 x 6}}{\text{.023 x 1}}{{\text{0}}^{ - 8}}{\text{ }}}}$
⟹The number of mole of electrons that weigh one kilogram =$\dfrac{1}{{{\text{9}}{\text{.108 x 6}}{\text{.023}}}}{\text{x 1}}{{\text{0}}^8}$
So, the correct option is A.
Additional information: A mole is the SI unit to measure the amount of substance. Avogadro number is defined as the number of atoms present in 12g of carbon-12. The value of Avogadro number is\[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\] molecules/atoms. It is denoted as \[{\text{\;}}{{\text{N}}_{\text{A}}}\]. Therefore, number of moles can also be calculated from the Avogadro number.
\[{\text{Number of moles = }}\dfrac{{{\text{Number of particles}}}}{{{\text{Avogadro Number}}}}\]
Note: Mass of an electron can also be expressed in g since 1kg = \[{\text{1}}{{\text{0}}^{\text{3}}}\]g therefore mass of electron becomes ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{g}}$ therefore proper unit conversion is necessary.
Complete step by step answer: It is given that mass of one electron = ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ kg}}$
We know that one mole of any substance has an Avogadro number of atoms or molecules or ions. Avogadro number,\[{\text{\;}}{{\text{N}}_{\text{A}}}\]= \[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\]
⟹Mass of one mole of electrons = ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ x 6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}$kg
Since in the options we can see that the numbers are not in simplified form so we do not perform the multiplication we can keep it as such till the end.
Now, we need to find the number of moles of electrons that weigh one kilogram(kg)
⟹ The number of mole of electrons that weigh one kilogram = $\dfrac{1}{{{\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 31}}}}{\text{ x 6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}}}$
Now let’s just simplify the power terms, by doing so we get,
⟹The number of mole of electrons that weigh one kilogram = $\dfrac{1}{{{\text{9}}{\text{.108 x 6}}{\text{.023 x 1}}{{\text{0}}^{ - 8}}{\text{ }}}}$
⟹The number of mole of electrons that weigh one kilogram =$\dfrac{1}{{{\text{9}}{\text{.108 x 6}}{\text{.023}}}}{\text{x 1}}{{\text{0}}^8}$
So, the correct option is A.
Additional information: A mole is the SI unit to measure the amount of substance. Avogadro number is defined as the number of atoms present in 12g of carbon-12. The value of Avogadro number is\[{\text{6}}{\text{.023 x 1}}{{\text{0}}^{{\text{23}}}}\] molecules/atoms. It is denoted as \[{\text{\;}}{{\text{N}}_{\text{A}}}\]. Therefore, number of moles can also be calculated from the Avogadro number.
\[{\text{Number of moles = }}\dfrac{{{\text{Number of particles}}}}{{{\text{Avogadro Number}}}}\]
Note: Mass of an electron can also be expressed in g since 1kg = \[{\text{1}}{{\text{0}}^{\text{3}}}\]g therefore mass of electron becomes ${\text{9}}{\text{.108 x 1}}{{\text{0}}^{{\text{ - 27}}}}{\text{g}}$ therefore proper unit conversion is necessary.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

