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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