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In a periodic table, the average mass of magnesium is given as 24.312 u. The average value of based on their relative natural abudance on earth. The three isotopes and their masses are ""_(12)^(24)Mg (23.98504 u), ""_(12)^(25)Mg(24.98584 u) and ""_(42)^(26)Mg(25.98259u). The natural abundance of ""_(12)^(24)Mg is 78.99% by mass. Calculate the abundances of ther two isotopes. |
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Answer» Solution :`(""_(12)^(24)Mg) = 24.312 u` Mass of `""_(12)^(24)Mg = 23.985044` `""_(12)^(25)Mg = 24.98584u` `""_(12)^(25)Mg = 24.98584u` `""_(12)^(26)Mg = 25.98259u` Let us assume that there are `n_1` atoms of `I^(ST)` isotope, `n_2` atoms of `2^(nd)` etc. TOTAL `(n_1 + n_2 + n_3)` HENCE `(n_1 + n_2 + n_3)24.312` GIVES the total mass. Relative abundance of first = `(n_1)/((n_1 + n_2 + n_3))` Relative abundance of `2^(nd) = (n^2)/((n_1 + n_2 + n_3))` Relative abdundance of `3^(rd) = (n_3)/((n_1 + n_2 + n_3))` `:. n_1 xx 23.98504 + n_2 xx 24.98584 + n_3 xx 25.98259 = (n_1 + n_2 + n_3)24.3124` so that `(n_1)/((n_1 + n_2 + n_3)) xx 23.98504 + (n_2)/((n_1 + n_2 + n_3)) xx 24.98594 + (n_3)/((n_1 + n_2 + n_3)) xx 25.98259` `= 24.3126` `(n_1)/((n_1 + n_2 + n_3)) = 0.7899, (n_2)/((n_1 + n_2 + n_3)) = x, (n_3)/((n_1 + n_2 + n_3)) = 0.2101 - x` `:. 0.7899 xx 23.98504 + x xx 24.98584 + 0.2101 xx 25.98259 - x xx 25.98259 - x xx 25.98259 = 24.3126` `18.9457515 + 5.458942153 - 24.3126 = (25.98259 - 24.98584)x` `x = (0.092093653)/(0.99675) = 0.09239` i.e, 9.24% Adundance of last isotope = 0.1177 i.e., 11.77 |
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