
Which of the following expressions is correct (n = no. of moles of the gas, ${{\text{N}}_{\text{A}}}$ = Avogadro constant, m is the mass of 1 molecule of the gas, N is the no. of molecules of the gas)?
A)\[\text{n = m}{{\text{N}}_{\text{A}}}\]
B) \[\text{m = }{{\text{N}}_{\text{A}}}\]
C) \[\text{N = n}{{\text{N}}_{\text{A}}}\]
D) \[\text{m =}\dfrac{\text{n}}{{{\text{N}}_{\text{A}}}}\]
Answer
576.3k+ views
Hint: Avogadro’s number ${{\text{N}}_{\text{A}}}$ is widely used to count the number of units present in the one mole of the substance. According to which the one mole of a substance contains the $6.022\text{ }\times \text{ }{{10}^{23}}$particles. Therefore if,
$\begin{align}
& \text{1 mole of substance= 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ partciles} \\
& \text{x mole of substance = }\dfrac{\text{Given number of particles }}{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ }} \\
\end{align}$
Complete step by step answer:
We know that atoms and molecules are microscopic particles. This cannot be seen by naked eyes. They are so small that they cannot be counted down. Therefore chemist uses the concept of mole for counting the atoms, molecules, or ions in the substance. The moles are represented by ‘n’.
We are also well familiar with Avogadro's number. It is a measure of the number of units present in one mole of substance which is equal to the \[6.02214076\text{ }\times \text{ }{{10}^{23}}\] particles. These units can be an electron, atoms, ions or molecules depend on the nature of the substance. In other words, one mole of a substance contains the \[6.02214076\text{ }\times \text{ }{{10}^{23}}\]particles.
Let us take an example for 1 mole of the atom.
We know that,
$1\text{ mole of atoms = Avagadro }\!\!'\!\!\text{ s Number = }6.022\text{ }\times \text{ }{{10}^{23}}\text{ atoms}$
Since the number of particles present in 1 mole of any substance is fixed.
Therefore the number of moles can be represented in terms of Avogadro's number is:
\[\text{Number of moles = }\dfrac{\text{Given number of atom or particle}}{\text{Avagadro }\!\!'\!\!\text{ s number}}\]
It can be also represented as follows,
\[\text{n = }\dfrac{\text{N}}{{{\text{N}}_{\text{A}}}}\] (1)
Where n is the number of moles, N is the given number of particles and \[{{\text{N}}_{\text{A}}}\]is the Avogadro's number.
The equation (1) can be interchanged as,
\[\begin{align}
& \text{n = }\dfrac{\text{N}}{{{\text{N}}_{\text{A}}}} \\
& \Rightarrow \text{N= n }\!\!\times\!\!\text{ }{{\text{N}}_{\text{A}}} \\
\end{align}\]
$\text{number of particles of element , N = n }\times \text{ }{{\text{N}}_{\text{A}}}$
Hence, (C) is the correct option.
Additional information:
Now we know that number of mole ‘n’ can be explained in terms of mass and molar mass. The relation is as follows:
\[\begin{align}
& \text{Number of moles = }\dfrac{\text{Mass of element}}{\text{Molar mass}} \\
& \text{ }OR \\
& \text{ n = }\dfrac{\text{m}}{\text{M}} \\
\end{align}\] (2)
Where n is the number of moles, m is the given mass and M is the molar mass. Thus we can rewrite the equation as
$\text{mass of element , m = n }\!\!\times\!\!\text{ M}$
Note: For volume at the STP condition the one mole is equalled to the $22.4\text{ litres}$ of the gas.
$\begin{align}
& \text{1 mole of substance= 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ partciles} \\
& \text{x mole of substance = }\dfrac{\text{Given number of particles }}{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ }} \\
\end{align}$
Complete step by step answer:
We know that atoms and molecules are microscopic particles. This cannot be seen by naked eyes. They are so small that they cannot be counted down. Therefore chemist uses the concept of mole for counting the atoms, molecules, or ions in the substance. The moles are represented by ‘n’.
We are also well familiar with Avogadro's number. It is a measure of the number of units present in one mole of substance which is equal to the \[6.02214076\text{ }\times \text{ }{{10}^{23}}\] particles. These units can be an electron, atoms, ions or molecules depend on the nature of the substance. In other words, one mole of a substance contains the \[6.02214076\text{ }\times \text{ }{{10}^{23}}\]particles.
Let us take an example for 1 mole of the atom.
We know that,
$1\text{ mole of atoms = Avagadro }\!\!'\!\!\text{ s Number = }6.022\text{ }\times \text{ }{{10}^{23}}\text{ atoms}$
Since the number of particles present in 1 mole of any substance is fixed.
Therefore the number of moles can be represented in terms of Avogadro's number is:
\[\text{Number of moles = }\dfrac{\text{Given number of atom or particle}}{\text{Avagadro }\!\!'\!\!\text{ s number}}\]
It can be also represented as follows,
\[\text{n = }\dfrac{\text{N}}{{{\text{N}}_{\text{A}}}}\] (1)
Where n is the number of moles, N is the given number of particles and \[{{\text{N}}_{\text{A}}}\]is the Avogadro's number.
The equation (1) can be interchanged as,
\[\begin{align}
& \text{n = }\dfrac{\text{N}}{{{\text{N}}_{\text{A}}}} \\
& \Rightarrow \text{N= n }\!\!\times\!\!\text{ }{{\text{N}}_{\text{A}}} \\
\end{align}\]
$\text{number of particles of element , N = n }\times \text{ }{{\text{N}}_{\text{A}}}$
Hence, (C) is the correct option.
Additional information:
Now we know that number of mole ‘n’ can be explained in terms of mass and molar mass. The relation is as follows:
\[\begin{align}
& \text{Number of moles = }\dfrac{\text{Mass of element}}{\text{Molar mass}} \\
& \text{ }OR \\
& \text{ n = }\dfrac{\text{m}}{\text{M}} \\
\end{align}\] (2)
Where n is the number of moles, m is the given mass and M is the molar mass. Thus we can rewrite the equation as
$\text{mass of element , m = n }\!\!\times\!\!\text{ M}$
Note: For volume at the STP condition the one mole is equalled to the $22.4\text{ litres}$ of the gas.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

