
\[3.0\times {{10}^{23}}\]atoms of neon weigh:
(A) 0.50 gram
(B) 1.0 gram
(C) 5.0 gram
(D) 40.0 gram
(E) 10.0 gram
Answer
233.4k+ views
Hint: To solve this question, we should first know about the atomic weight of neon. We should note that the atom present in one mole of substance is equal to Avogadro’s number.
Complete step by step answer:
We should first know about one important concept. This concept will be helpful for us to solve this question. In this concept we will learn about Avogadro numbers. We should know that Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to\[6.022\times {{10}^{23}}\]\[mol{{e}^{\,-1}}\]and is expressed as the symbol NA. Any by using this concept we can say that one mole of any element is equal to particles equal to Avogadro number.
As we know that the mass of one mole of substance is called the molar mass of that substance.
The molar mass of neon is 20 gram per mole.
From the above discussion, we know that one mole of substance has =\[6.022\times {{10}^{23}}\]atoms (Avogadro's number).
The number of moles in \[3.0\times {{10}^{23}}\]atoms =\[\dfrac{3.0\times {{10}^{23}}}{6.02\times {{10}^{23}}~}=0.5\,mole\]
Weight of one mole of neon is 20 gram per mole.
So, weight of 0.5 mole of neon is equal to: \[0.5\times 20=10\]grams
So, from the above discussion and calculation, we find out that our correct answer is option E.
Note: We should clear our concept between amu and gram per mole. For our knowledge let us now discuss the difference between amu and gram per mole. We know that each ion, or atom, has a definite mass; similarly, each mole of a given pure substance also has a definite mass. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units (amu) or in grams per mole\[(\dfrac{g}{mole}).\] So, from this discussion we now know that we can express mass in any of the units.
Complete step by step answer:
We should first know about one important concept. This concept will be helpful for us to solve this question. In this concept we will learn about Avogadro numbers. We should know that Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to\[6.022\times {{10}^{23}}\]\[mol{{e}^{\,-1}}\]and is expressed as the symbol NA. Any by using this concept we can say that one mole of any element is equal to particles equal to Avogadro number.
As we know that the mass of one mole of substance is called the molar mass of that substance.
The molar mass of neon is 20 gram per mole.
From the above discussion, we know that one mole of substance has =\[6.022\times {{10}^{23}}\]atoms (Avogadro's number).
The number of moles in \[3.0\times {{10}^{23}}\]atoms =\[\dfrac{3.0\times {{10}^{23}}}{6.02\times {{10}^{23}}~}=0.5\,mole\]
Weight of one mole of neon is 20 gram per mole.
So, weight of 0.5 mole of neon is equal to: \[0.5\times 20=10\]grams
So, from the above discussion and calculation, we find out that our correct answer is option E.
Note: We should clear our concept between amu and gram per mole. For our knowledge let us now discuss the difference between amu and gram per mole. We know that each ion, or atom, has a definite mass; similarly, each mole of a given pure substance also has a definite mass. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units (amu) or in grams per mole\[(\dfrac{g}{mole}).\] So, from this discussion we now know that we can express mass in any of the units.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

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

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

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

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

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

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

