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danny_007

A fused mixture of rubidium fluoride and uranium (IV) fluoride can be oxidized with fluorine to produce a uranium compound in which uranium is mainly bt not entirely in +5 oxidation state , the product is found to contact 54.93% of uranium. A 1.0357 g sample of product is immersed in 100 ml of 10.07 centimolar acidified KI solution

2 I^- + 2UF₆^- gives 2UF₄ + I₂ + 4F^-

the iodine produces was titrated with 14.80 ml of 14.94 centimolar sodium thiosulphate sol . what % of the original mixture oxidezed to +5 oxidation state?

danny_007

......................................?

danny_007

chimanshu_007

93% approx.....rite?

danny_007

dev_22oct

2I^- + 2UF₆^- -------> 2UF₄ + I₂+ 4F^- (1)

I₂ + 2S₂O₃^2- ------> 2I^-+ S₄O₆^2- (B)

thus, molar ratio :: 2S₂O₃^2- = I₂ = 2UF₆^-

14.8 mL of 0.1494/100 M Na2S2O3 soln = 14.8 X 0.1494 milliomoles

= millimoles of UF6-
= 14.8 X 0.1494 X 10^-3 X 238 gm U (in +5 Os) = K Suppose ------ (i)

now, wt of Uranium in 1.037 gm product = 0.5443 X 1.0357 ------(II)

thus, % of Uranium oxidi$ed to +5 O$
= (I)/(II)

= 93 % approX

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