Answer:
Abundance of Iridium-193 is 62.75%
Explanation:
From the question given above, the following data were obtained:
Isotope A (Iridium-191):
Mass of A = 190.9605 amu
Abundance of A = A%
Isotope B (Iridium-193):
Mass of B = 192.9629 amu
Abundance B = (100 – A) %
Relative atomic mass of Iridium = 192.217 amu
Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:
Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]
192.217 = 1.909605A% + 1.929629(100 – A)%
192.217 = 1.909605A% + 192.9629 – 1.929629A%
Collect like terms
192.217 – 192.9629 = 1.909605A% – 1.929629A%
–0.7459 = –0.020024A%
Divide both side by –0.020024
A% = –0.7459 / –0.020024
A% = 37.25 %
Finally, we shall determine the abundance of Isotope B (Iridium-193).
This can be obtained as follow:
Abundance of A (Iridium-191) = 37.25 %
Abundance of B (Iridium-193) =?
Abundance B = 100 – A%
Abundance B = 100 – 37.25 %
Abundance of B (Iridium-193) = 62.75%
