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Uranium is isolated from its ore by dissolving it as `UO_(2)(NO_(3))_(2)` and separating it as solid `UO_(2)(C_(2)O_(4)).xH_(2)O " "A 1.0 g` sample of ore on treatment with nitric acid yielded `1.48 g UO_(2)(NO_(3))_(2)` which on further treatment with `0.4 g Na_(2)C_(2)O_(4)` yielded `1.23 g " "UO_(2)" "(C_(2)O_(4)).xH_(2)O`. Determine weight percentage of uranium in the original sample and `x` : |
Answer» Correct Answer - `89.4.3` Mass of uranium in the sample `=(1.48)/(394)xx238=0.894g` Mass `%` of uranium in the sample `=89.4` `UO_(2)(NO_(3))_(2)+Na_(2)C_(2)O_(4)+xH_(2)OrarrUO_(2)(C_(2)O_(4))xH_(2)Odarr` `m" "mol 3.756" "2.985" "+2NaNO_(3)` Here `Na_(2)C_(2)O_(4)` is the limiting reagent, therefore, `m` mol of `UO_(2)(C_(2)O_(4)).xH_(2)O` formed is `2.985` `implies M(UO_(2))(C_(2) O_(4)).xH_(2)O=(1.23)/(2.985)xx1000=412` `238+32+88+18x` `implies x =(54)/(18) = 3` |
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