Question

A natural sample of Antimony gas consist Sb121 and Sb123 isotope. If the average atomic mass of Sb is 121.5 g. Select the option regarding the percentage abundance of Sb121 and Sb123 respectively.


80%, 20%


75%, 25%


40%, 60%


50%, 50%

Answer 1

Answer:

80%57% it's answer is that

40%20%

Answer 2

Given:

The average atomic mass of Sb = 121.5 gm

To Find:

The percentage abundance of Sb121 and Sb123 isotopes.

Calculation:

- Let the percentage abundance of Sb121 be x %.

⇒ The percentage abundance of Sb123 will be (100-x) %.

- The average atomic mass is given as:

Average = M1P1 + M2P2

⇒ 121.5 = 121 × x/100 + 123 × (100-x)/100

⇒ 121.5 = 1.21x + 123 - 1.23x

⇒  0.02x = 1.5

⇒  x = 1.5/0.02

⇒  x = 75 %

- The abundance of Sb121, i.e., x = 75 %

- The abundance of Sb123, i.e., (100-x) % = 100-75 = 25 %

-So, the correct answer is option (b) 75 %, 25 %