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The element [tex]$X$[/tex] has three naturally occurring isotopes. The isotopic masses (amu) and \% abundances of the isotopes are given in the table below. The average atomic mass of the element is \_\_\_\_\_ amu.

\begin{tabular}{|c|c|c|}
\hline Isotope & Abundance (\%) & Mass (amu) \\
\hline 53X & 19.61 & 52.62 \\
\hline 56X & 53.91 & 56.29 \\
\hline 58X & 26.48 & 58.31 \\
\hline
\end{tabular}

A. 33.33
B. 55.74
C. 56.11


Sagot :

To determine the average atomic mass of element [tex]\(X\)[/tex] given the isotopic masses and their respective abundances, we follow these steps:

1. Convert the percentage abundances to decimal form:
- For isotope 53x: [tex]\(19.61\% = \frac{19.61}{100} = 0.1961\)[/tex]
- For isotope 56x: [tex]\(53.91\% = \frac{53.91}{100} = 0.5391\)[/tex]
- For isotope 58x: [tex]\(26.48\% = \frac{26.48}{100} = 0.2648\)[/tex]

2. Multiply each isotope's mass by its respective decimal abundance:
- For isotope 53x: [tex]\(0.1961 \times 52.62 = 10.307082\)[/tex]
- For isotope 56x: [tex]\(0.5391 \times 56.29 = 30.353439\)[/tex]
- For isotope 58x: [tex]\(0.2648 \times 58.31 = 15.444688\)[/tex]

3. Sum the results from step 2 to find the average atomic mass:
[tex]\[ 10.307082 + 30.353439 + 15.444688 = 56.105209 \][/tex]

Thus, the average atomic mass of element [tex]\(X\)[/tex] is approximately [tex]\(56.11 \, \text{amu}\)[/tex].

So the final answer to the question is:
[tex]\[ \boxed{56.11} \][/tex]
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