Respuesta :
Option C: 106.91 amu
Atomic mass of silver is 107.868 amu. Mass of Ag-109 isotope is 108.91 amu and its percentage abundance is 48.16%, let the mass of other isotope be X, the percentage abundance can be calculated as:
%X=100-48.16=51.84%
Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.
Thus,
Atomic mass= m(Ag-109)×%(Ag-109)+X×%(Ag-X)
Putting the values,
[tex]107.868 amu=(108.91 amu)(\frac{48.16}{100} )+X(\frac{51.84}{100} )[/tex]
[tex]107.868 amu=52.45 amu+X(\frac{51.84}{100} )[/tex]
Thus,
[tex]X=\frac{107.868-52.45}{0.5184} =106.9[/tex]
Therefore, other isotope is Ag-106.9, which is close to option C.
The mass, in amu, of the other isotope of the silver is 106.91 amu
Let the 1st isotope be A
Let the 2nd isotope be B
From the question given above, the following data were obtained:
Atomic mass of Ag = 107.868 amu
For isotope A:
Mass of A = 108.91 amu
Abundance of A (A%) = 48.16%
For isotope B:
Abundance of B (B%) = 100 – 48.16 = 51.84%
Mass of A = ?
The mass of the other isotope can be obtained as follow:
Atomic mass = [(mass of A × A%)/100] + [(mass of B × B%)/100]
107.868 = [(108.91 × 48.16)/100] + [(mass of B × 51.84)/100]
107.868 = 52.451056 + mass of B × 0.5184
Collect like terms
107.868 – 52.451056 = mass of B × 0.5184
55.416944 = mass of B × 0.5184
Divide both side by 0.5184
Mass of B = 55.416944 / 0.5184
Mass of B = 106.9 amu
Therefore, the mass of the other isotope is 106.91 amu
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