Given the average atomic weight of Chlorine is 35.45, and the masses of its isotopes are 34.96 for 35 extCl and 36.98 for 37 extCl, calculate the relative abundance of 35 extCl?

Let $x$ be the relative abundance of $^{35}\text{Cl}$.
Since there are only two isotopes, the relative abundance of $^{37}\text{Cl}$ will be $(1-x)$.

The average atomic weight of an element is calculated by summing the product of each isotope's mass and its relative abundance.
Average atomic weight = $(\text{mass of } ^{35}\text{Cl} \times \text{abundance of } ^{35}\text{Cl}) + (\text{mass of } ^{37}\text{Cl} \times \text{abundance of } ^{37}\text{Cl})$
Given:

• Average atomic weight of Chlorine = 35.45
• Mass of $^{35}\text{Cl}$ = 34.96
• Mass of $^{37}\text{Cl}$ = 36.98

Substitute these values into the equation:

Combine the terms with $x$:

Subtract 36.98 from both sides:

Divide by $-2.02$:

To convert the decimal abundance to a percentage, multiply by $100\%$:

The relative abundance of $^{35}\text{Cl}$ is approximately $\boxed{\text{75.74\%}}$.