Question

Vitellium (Vi) has the following composition:

Answer 1

Answer:

The average atomic mass of vitellium is 192.11 amu.

Explanation:

The average atomic mass is an average between the masses of all the isotopes, considering their abundance in the nature. It can be calculated using the following expression:

$AAM=\frac{\Sigma m_{i} \times ab_{i} }{100}$

where,

$m_{i}$ is the mass of each isotope.

$ab_{i}$ is the abundance in the nature of each isotope.

For vitelium,

$AAM=\frac{187.9122amu \times 10.861 + 190.9047amu \times 12.428 + 192.8938 amu\times 76.711 }{100} =192.11amu$

Answer 2

192.1055 to 4 decimal places. If you want the workings out, don't hesitate to ask.