Question

Natural chlorine consists of ${}^{35}Cl$ and ${}^{37}Cl$ . Their atomic masses are 34.97 amu and 36.97 amu respectively. If the average mass of chlorine is 35.5 amu. Calculate the abundance of ${}^{35}Cl$ is :
a.) 24.0%
b.) 35.5%
c.) 50.0%
d.) 75.5%

Answer 1

Hint: In order to deal with this question we will give a brief explanation about chlorine, further then we will assume abundance of ${}^{35}Cl$ as a variable and we will apply the formula of average atomic mass to get the answer.

Complete step-by-step answer:
Chlorine is a chemical element with the symbol $Cl$ and atomic number 17. The second-lightest of the halogens, it occurs in the periodic table between fluorine and bromine, and its properties are often intermediate among them. At room temperature chlorine is a yellow-green gas. It is an incredibly reactive element and a heavy oxidizing agent: it has the highest electron affinity of the elements and the third highest electronegativity on the Pauling scale, behind oxygen and fluorine only.
Now come to the question:
As we know that the percent abundance of ${}^{35}Cl$ and ${}^{37}Cl$ is equal to 100.
Let us assume abundance of ${}^{35}Cl = x$ and abundance of ${}^{37}Cl = 100 - x$
Given that :
Atomic weight of ${}^{35}Cl = 34.97amu$
Atomic weight of ${}^{37}Cl = 36.97amu$
Atomic mass of $Cl = 35.5amu$
We know the formula of average atomic mass:
Average atomic mass = sum of products of average atomic mass for each isotope and their percentage abundance.
So, Average atomic mass
$\Rightarrow \dfrac{{\left( {34.97 \times x} \right) + \left( {36.97 \times \left( {100 - x} \right)} \right)}}{{100}} = 35.5$
Further simplify the above equation we get

$$\Rightarrow \dfrac{{3697 - \left( {36.97x - 34.97x} \right)}}{{100}} = 35.5 \\\ \Rightarrow \dfrac{{3697 - \left( {2x} \right)}}{{100}} = 35.5 \\\ \Rightarrow 2x = 3697 - 3550 \\\ \Rightarrow 2x = 147 \\\ \Rightarrow x = 73.5\% \\\$$

x = 73.5 %
Hence, Abundance of ${}^{35}Cl$ is around 75.5%
So, the correct answer is option D.

Note- Chlorine has two stable isotopes, ${}^{35}Cl$ and ${}^{37}Cl$. These are its only two natural isotopes occurring in quantity, with ${}^{35}Cl$ making up 76% of natural chlorine and ${}^{37}Cl$ making up the remaining 24%. The oxygen-burning and silicon-burning mechanisms are also synthesized in stars. They have nuclear spin $\dfrac{3}{2} +$ and can thus be used for nuclear magnetic resonance, however the spin magnitude is greater than $\dfrac{1}{2}$, resulting in non-spherical distribution of nuclear charges and thus resonance expansion as a result of a non-zero quadrupolar moment and consequent quadrupolar relaxation.