Answer
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Hint : We will solve this problem with the help of the formula of finding the average relative atomic mass of an atom when atomic masses of their isotopes are known. Let suppose any atom has two isotopes then the formula of average relative atomic mass of atom is; Average relative atomic mass of a atom= [(atomic mass of first isotope× percentage of abundance) +(atomic mass of second isotope× percentage of abundance)]/$100$. We know that isotopes are defined as the species that are one of two or more species of atoms of an element which have the same atomic number and position in the periodic table but different atomic masses.
Complete step by step solution:
Let take percentage abundance of $^{17}{C_{35}}$ is $x\% $ then percentage of abundance of $^{17}{C_{37}}$ will be $(100 - x)\% $ , now with the help of above formula we can calculate the value of $x$.
Average relative atomic mass= [(atomic mass of first isotope× percentage of abundance) +(atomic mass of second isotope× percentage of abundance)]/$100$
$\begin{gathered}
\Rightarrow 35.5 = \dfrac{{35x + 37(100 - x)}}{{100}} \\
\Rightarrow 3550 = 35x + 3700 - 3550 \\
\Rightarrow 2x = 150 \\
\Rightarrow x = 75 \\
\end{gathered} $
Hence the percentage of heavier isotope which is $^{17}{C_{37}}$ will be $100 - 75 = 25\% $. So here option D is the correct answer to this question.
Note: We know that the chlorine atom has two isotopes of atomic mass $35$ and $37$, we have calculated that their abundance is $75\% $ and $25\% $ respectively. Because chlorine has $^{17}{C_{35}}$ and $^{17}{C_{37}}$ , two isotopes hence average of their mass is $35.5$ which is used as atomic mass of chlorine. Out of four Chlorine atoms there are three $^{17}{C_{35}}$, therefore we take the average of all chlorine atoms. These two are the most stable isotopes of chlorine atoms.
Complete step by step solution:
Let take percentage abundance of $^{17}{C_{35}}$ is $x\% $ then percentage of abundance of $^{17}{C_{37}}$ will be $(100 - x)\% $ , now with the help of above formula we can calculate the value of $x$.
Average relative atomic mass= [(atomic mass of first isotope× percentage of abundance) +(atomic mass of second isotope× percentage of abundance)]/$100$
$\begin{gathered}
\Rightarrow 35.5 = \dfrac{{35x + 37(100 - x)}}{{100}} \\
\Rightarrow 3550 = 35x + 3700 - 3550 \\
\Rightarrow 2x = 150 \\
\Rightarrow x = 75 \\
\end{gathered} $
Hence the percentage of heavier isotope which is $^{17}{C_{37}}$ will be $100 - 75 = 25\% $. So here option D is the correct answer to this question.
Note: We know that the chlorine atom has two isotopes of atomic mass $35$ and $37$, we have calculated that their abundance is $75\% $ and $25\% $ respectively. Because chlorine has $^{17}{C_{35}}$ and $^{17}{C_{37}}$ , two isotopes hence average of their mass is $35.5$ which is used as atomic mass of chlorine. Out of four Chlorine atoms there are three $^{17}{C_{35}}$, therefore we take the average of all chlorine atoms. These two are the most stable isotopes of chlorine atoms.
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