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In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundnce on earth. The three isotopes and their masses are `._(12)^(24)Mg (23.98504 u), ._(12)^(25)Mg (24.98584 u)` and `._(12)^(26)Mg(25.98259 u)`. The natural abundance of `._(12)^(24)Mg` is `78.99 %` by mass. Calculate the abundances of other two isotopes. |
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Answer» Let the abundance of `._(12)Mg^(25)` by mass be x% therefore, abundance of `._(12)Mg^(26)` by mass = (100 - 78.99) - x% = (21.01) - x% Now average atomic mass of magnesium is `24.312 = (23.98504xx78.99+24.98584+25.98529 (21.01 - x))/(100)` on solving we get x = 9.303 % for `._(12)Mg^(25)` and for `._(12)Mg^(26), (21.01 -x) = 11.71 %` |
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