Naturally occurring argon (${\text{Ar}}$) consists of three isotopes the atom of which occur in the abundances as given in the table. The atomic weight of argon is:
\[\begin{gathered}
{\text{Isotopes}} \\
{}^{{\text{36}}}{\text{Ar}} \\
{}^{{\text{38}}}{\text{Ar}} \\
{}^{{\text{40}}}{\text{Ar}} \\
\end{gathered} \] $\begin{gathered}
{\text{Abundance}} \\
0.34\% \\
0.07\% \\
99.59\% \\
\end{gathered} $
(A)- ${\text{38}}{\text{.45u}}$
(B)- ${\text{37}}{\text{.20u}}$
(C)- ${\text{39}}{\text{.98u}}$
(D)- ${\text{40}}{\text{.98u}}$
Answer
598.5k+ views
Hint: Whenever in the question isotopes are given then in this condition atomic mass of atom is calculated by taking the average weight of all isotopes i.e. by adding atomic weight of all isotopes which is divided by all total covered volume of isotopes.
Complete step by step solution: Isotopes are that species of given element in which no. of electrons or protons are equal but neutrons are not equal i.e. atomic masses are different. In the question the percentage & abundance of available isotopes are given where,
Atomic mass of ${}^{{\text{36}}}{\text{Ar = 36}}$,
Atomic mass of ${}^{{\text{38}}}{\text{Ar = 38}}$,
Atomic mass of ${}^{40}{\text{Ar = 40}}$.
For calculating atomic mass of argon we have to add atomic mass of all given isotopes & calculation is as follow:
Atomic mass $ = \dfrac{{\left( {0.34 \times 36} \right) + \left( {0.07 \times 38} \right) + \left( {99.59 \times 40} \right)}}{{100}}$
Atomic mass $ = \dfrac{{12.24 + 2.66 + 3983.6}}{{100}}$
Atomic mass ${\text{ = 39}}{\text{.98u}}$
Hence, option (C) is correct.
Additional information: In the calculation part we multiplied the atomic masses by 0.34, 0.07 & 99.59 because we had to add the atomic mass of all isotopes which are present in abundance or in a given quantity, and 36, 38 & 40 are atomic mass of single atom only.
Note: Here some of you may think that why we divide the sum of all atomic masses by 100, so the reason behind this is that the total covered percentage of given isotopes is 100 that’s why we divide the sum by 100.
Complete step by step solution: Isotopes are that species of given element in which no. of electrons or protons are equal but neutrons are not equal i.e. atomic masses are different. In the question the percentage & abundance of available isotopes are given where,
Atomic mass of ${}^{{\text{36}}}{\text{Ar = 36}}$,
Atomic mass of ${}^{{\text{38}}}{\text{Ar = 38}}$,
Atomic mass of ${}^{40}{\text{Ar = 40}}$.
For calculating atomic mass of argon we have to add atomic mass of all given isotopes & calculation is as follow:
Atomic mass $ = \dfrac{{\left( {0.34 \times 36} \right) + \left( {0.07 \times 38} \right) + \left( {99.59 \times 40} \right)}}{{100}}$
Atomic mass $ = \dfrac{{12.24 + 2.66 + 3983.6}}{{100}}$
Atomic mass ${\text{ = 39}}{\text{.98u}}$
Hence, option (C) is correct.
Additional information: In the calculation part we multiplied the atomic masses by 0.34, 0.07 & 99.59 because we had to add the atomic mass of all isotopes which are present in abundance or in a given quantity, and 36, 38 & 40 are atomic mass of single atom only.
Note: Here some of you may think that why we divide the sum of all atomic masses by 100, so the reason behind this is that the total covered percentage of given isotopes is 100 that’s why we divide the sum by 100.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

