Urnaium is isolated from its ore by dissolving it as `UO_2(NO_3)_2` and separating it as solid `UO_2(C_2O_4).x H_2O`A 1.0 g sample of ore on treatment
Urnaium is isolated from its ore by dissolving it as `UO_2(NO_3)_2` and separating it as solid `UO_2(C_2O_4).x H_2O`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_2C_2O_4` yielded 1.23 g`UO_2(C_2O_4).xH_2O`.Determine weight percentage of uranium in the original sample of x.
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Correct Answer - 3
Mass of uranium in the sample `1.48/394xx238=0.894 g`
Now, `uderset(3.756)(UO_2(NO_3)_2)+underset(2.985)(Na_2C_2O_4)+xH_2Oto(UO_2(C_2O_4)xH_2Odarr + 2NaNO_3`
Here `Na_2C_2O_4` is the limiting reagent, therefore , m mole of `UO_2(C_2O_4).xH_2O` formed is 2.985.
`implies M(UO_2(C_2O_4)).xH_2O=1.23/2.985xx1000`=412=238+32+88+18x
`implies x=54/18=3`
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