Answer:
Mass of 22-Na contained in the sample = 0.0599 g
Explanation:
Mass of the isotope mixture = 1.8385g
Isotope mixture has apparent mass of 22.9573 u
23-Na has a relative atomic mass of 22.9898 u
22-Na has a relative atomic mass of 21.9944 u
Let the relative abundance of 23-Na be X
Then the relative abundance of 22-Na would be (1-X)
21.9944 (1-X) + 22.9898 X = 22.9573
21.9944 - 21.9944X + 22.9898X = 22.9573
22.9898X - 21.9944X = 22.9573 - 21.9944
0.9954X = 0.9639
X = 0.9674
Relative abundance of 23-Na = 0.9674
Relative abundance of 22-Na = 1 - 0.9674 = 0.0326
Mass of 22-Na in the 1.8385g of sample is
Relative abundance of 22-Na × Mass of sample = 0.0326 × 1.8385g = 0.0599 g
