Answer:
The mass of 22-Na included in the sample amounts to 0.0599 g
Explanation:
The total mass of the isotope mixture is 1.8385g.
It has an apparent mass of 22.9573 u.
For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.
Let the relative abundance of 23-Na be denoted as X.
This means that the relative abundance of 22-Na can be expressed as (1-X).
The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.
Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.
Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.
Hence, 0.9954X = 0.9639, leading to X = 0.9674.
The relative abundance of 23-Na is now identified as 0.9674.
Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.
Now, the mass of 22-Na contained within the 1.8385g sample is determined by
Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.
