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

`89.4.3`

Explanation

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`