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_2OA 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 gUO_2(C_2O_4).xH_2O.Determine weight percentage of uranium in the original sample of x.

Urnaium is isolated from its ore by dissolving it as UO_2(NO_3)_2 and

1.	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_2OA 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 gUO_2(C_2O_4).xH_2O.Determine weight percentage of uranium in the original sample of x.
Answer» Solution :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`