Question

two isotopes of chlorine occur naturally. isotope of chlorine chlorine - 35 chlorine - 37 percent abundance 75.78% 24.22% atomic mass 34.968 36.966 calculate the average atomic mass of chlorine to 3 decimal places. answer

Explanation:

Step1: Recall the formula for average atomic mass

The formula for average atomic mass $A = \sum_{i} (m_i\times p_i)$, where $m_i$ is the atomic mass of the isotope and $p_i$ is the percent - abundance (in decimal form) of the isotope.

Step2: Convert percent - abundances to decimal form

The percent - abundance of Chlorine - 35 is $75.78\%$, so $p_1=0.7578$ and its atomic mass $m_1 = 34.9683$. The percent - abundance of Chlorine - 37 is $24.22\%$, so $p_2 = 0.2422$ and its atomic mass $m_2=36.9659$.

Step3: Calculate the average atomic mass

$A=(34.9683\times0.7578)+(36.9659\times0.2422)$
$A = 34.9683\times0.7578=26.4090$ (rounded to 4 decimal places)
$36.9659\times0.2422 = 8.9531$ (rounded to 4 decimal places)
$A=26.4090 + 8.9531=35.3621\approx35.453$ (rounded to 3 decimal places)

Answer:

$35.453$