Question

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:

{\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 1

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.