Urnaium is isolated from its ore by dissolving it as $$UO2(NO3)2$$ and separating it as solid $$UO2(C2O4)$$.$$xH2O$$. A $$1.0$$ g sample of ore on treatment with nitric acid yielded $$1.48$$ g $$UO2(NO3)2$$ which on further treatment with $$0.4$$ g $$Na2C2O4$$ yielded $$1.23$$ g $$UO2(C2O4)$$. $$xH2O$$. Determine weight percentage of uranium in the original sample (y) and x. [At.wt of U $$=$$ $$238$$, Mw of $$UO2(NO3)2=394$$, Mw of $$UO2(C2O4)=358$$] Report (y/x)

Question

Urnaium is isolated from its ore by dissolving it as $$UO2(NO3)2$$ and separating it as solid $$UO2(C2O4)$$.$$xH2O$$. A $$1.0$$ g sample of ore on treatment with nitric acid yielded $$1.48$$ g $$UO2(NO3)2$$ which on further treatment with $$0.4$$ g $$Na2C2O4$$ yielded $$1.23$$ g $$UO2(C2O4)$$. $$xH2O$$. Determine weight percentage of uranium in the original sample (y) and x. [At.wt of U $$=$$ $$238$$, Mw of $$UO2(NO3)2=394$$, Mw of $$UO2(C2O4)=358$$] Report (y/x)

Infinity Learn · Accepted Answer

Mass of uranium in the sample $$1.48394$$×$$238=0.894g$$.Mass % of uranium in the sample $$=$$ $$89.4Now$$, $$UO2(NO3)2+Na2C2O4+xH2O$$→ $$UO2(C2O4)xH2O$$↓$$+2NaNO3$$.mmol                                $$3.756$$                                                $$2.985Here$$, $$Na2C2O4$$ is the limiting reagent, therefore, m mol of $$UO2(C2O4)$$.$$xH2O$$ formed is $$2.985$$.⇒$$M(UO2(C2O4))$$.$$xH2O=1.232.985$$×$$1000=412=238+32+88+18x$$.$$x=5418=3$$.

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