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Element [tex]$X$[/tex] has two isotopes. The table gives information about these isotopes.

\begin{tabular}{|c|c|c|}
\hline Isotope & \begin{tabular}{c}
Atomic \\
Mass (amu)
\end{tabular} & \begin{tabular}{c}
Abundance \\
(\%)
\end{tabular} \\
\hline [tex]$X -63$[/tex] & 62.9296 & 69.15 \\
\hline [tex]$X -65$[/tex] & 64.9278 & 30.85 \\
\hline
\end{tabular}

The average atomic mass of element [tex]$X$[/tex] is [tex]$\square$[/tex] amu.


Sagot :

To calculate the average atomic mass of element [tex]\( X \)[/tex], we need to use the given data for its isotopes. We will calculate the weighted mass contribution from each isotope and then sum these contributions.

1. Isotope [tex]\( X-63 \)[/tex]:
- Atomic mass: 62.9296 amu
- Abundance: 69.15%

2. Isotope [tex]\( X-65 \)[/tex]:
- Atomic mass: 64.9278 amu
- Abundance: 30.85%

Step 1: Calculate the weighted mass contribution of each isotope.

For isotope [tex]\( X-63 \)[/tex]:
[tex]\[ \text{Weighted mass of } X-63 = \text{Atomic mass of } X-63 \times \left(\frac{\text{Abundance of } X-63}{100}\right) \][/tex]
[tex]\[ \text{Weighted mass of } X-63 = 62.9296 \times \left(\frac{69.15}{100}\right) = 43.5158184 \, \text{amu} \][/tex]

For isotope [tex]\( X-65 \)[/tex]:
[tex]\[ \text{Weighted mass of } X-65 = \text{Atomic mass of } X-65 \times \left(\frac{\text{Abundance of } X-65}{100}\right) \][/tex]
[tex]\[ \text{Weighted mass of } X-65 = 64.9278 \times \left(\frac{30.85}{100}\right) = 20.0302263 \, \text{amu} \][/tex]

Step 2: Add the weighted mass contributions to determine the average atomic mass.

[tex]\[ \text{Average atomic mass} = \text{Weighted mass of } X-63 + \text{Weighted mass of } X-65 \][/tex]
[tex]\[ \text{Average atomic mass} = 43.5158184 \, \text{amu} + 20.0302263 \, \text{amu} = 63.5460447 \, \text{amu} \][/tex]

Step 3: Round the average atomic mass to the nearest hundredth.

[tex]\[ 63.5460447 \, \text{amu} \approx 63.55 \, \text{amu} \][/tex]

Therefore, the average atomic mass of element [tex]\( X \)[/tex] is [tex]\( \boxed{63.55} \, \text{amu} \)[/tex].