A. The abundance of the 2nd isotope is 48.119%
B. The mass of the 2nd isotope is 108.905 amu
Let the 1st isotope be A
Let the 2nd isotope be B
A. Determination of the abundance of the 2nd isotope
Abundance of isotope A = 51.881%.
Abundance of B = 100 – A
Abundance of B = 100 – 51.881
B. Determination of the mass of the 2nd isotope
Atomic mass of silver = 107.868 amu.
Mass of 1st isotope (A) = 106.906 amu
Abundance of isotope A (A%) = 51.881%.
Abundance of isotope B (B%) = 48.119%
[tex]atomic \: mass = \frac{mass \: of \:A \times \:A\%}{100} + \frac{mass \: of \:B \times \:B\%}{100} \\ \\ 107.868 = \frac{106.906\times \ \: 51.881}{100} + \frac{mass \: of \:B \times \:48.119}{100} \\ \\ 107.868 = \: 55.464 + 0.48119 \times mass \: of \:B \\ \\ collect \: like \: terms \\ \\ 0.48119 \times mass \: of \:B = 107.868 - 55.464 \\ \\ divide \: both \: side \: by \: 0.48119 \\ \\ mass \: of \:B = \frac{107.868 - 55.464 }{0.48119} \\ \\ mass \: of \:B =108.905 \: amu \\ \\ [/tex]
Therefore, the mass of the 2nd isotope is 108.905 amu
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