Question

given the data in the table, what is the average atomic mass of chlorine?

	chlorine-35	chlorine-37
r dative abundance	75.78 percent	24.22 percent

a. 34.969 amu
b. 35.453 amu
c. 35.9675 amu
d. 36.966 amu

Explanation:

Step1: Recall the formula for average atomic mass

The formula for average atomic mass is the sum of (atomic mass of each isotope × its relative abundance (in decimal form)). So, we first convert the percentages to decimals. For Chlorine - 35, relative abundance in decimal is \( \frac{75.78}{100}=0.7578 \), and for Chlorine - 37, it is \( \frac{24.22}{100} = 0.2422 \).

Step2: Calculate the contribution of each isotope

For Chlorine - 35: \( 34.969\space amu\times0.7578 \). Let's calculate that: \( 34.969\times0.7578\approx34.969\times0.75 = 26.22675\) and \( 34.969\times0.0078\approx0.2727582 \), so total \( \approx26.22675 + 0.2727582=26.4995\space amu \) (approximate, more accurately \( 34.969\times0.7578 = 34.969\times(0.7 + 0.05+0.007 + 0.0008)=34.969\times0.7=24.4783\), \( 34.969\times0.05 = 1.74845\), \( 34.969\times0.007 = 0.244783\), \( 34.969\times0.0008 = 0.0279752\); sum: \( 24.4783+1.74845 = 26.22675+0.244783 = 26.471533+0.0279752 = 26.4995082\space amu \)).

For Chlorine - 37: \( 36.966\space amu\times0.2422 \). Calculate: \( 36.966\times0.2 = 7.3932\), \( 36.966\times0.04 = 1.47864\), \( 36.966\times0.0022 = 0.0813252 \). Sum: \( 7.3932+1.47864 = 8.87184+0.0813252 = 8.9531652\space amu \).

Step3: Sum the contributions

Now, add the two contributions: \( 26.4995082+8.9531652 = 35.4526734\space amu\approx35.453\space amu \).

Answer:

B. 35.453 amu