A chemist analyses samples containing no more than 0.476 g of uranium, prepared as U 4 + and oxidized by Mn O 4 − in acid: U 4 + + Mn O 4 − + H 2 O → U O 2 2 + + M n 2 + + H 3 O + . If the t

Question

A chemist analyses samples containing no more than 0.476 g of uranium, prepared as $U^{4 +}$ and oxidized by $Mn O_{4}^{-}$ in acid: $U^{4 +}$ + $Mn O_{4}^{-}$ + $H_{2} O$ $\to$ $U O_{2}^{2 +}$ + $M n^{2 +}$ + $H_{3} O^{+}$ . If the total $U^{4 +}$ sample must react with a maximum of 25 ml of $KMn O_{4}$ solution, what concentration does he choose? (U = 238)

Options:

Answer:

B

Solution:

mol U = 0.476/238 = 2 mmol; 5 $U^{4 +}$ : 2 $Mn O_{4}^{-}$ , so $KMn O_{4}$ = 2 $x$ 2/5 = 0.8 mmol. Concentration = 0.8/25 = 0.032 M, option (b).

TestHub · Accepted Answer

mol U = 0.476/238 = 2 mmol; 5 U 4 + : 2 Mn O 4 − ​ , so KMn O 4 ​ = 2 x 2/5 = 0.8 mmol. Concentration = 0.8/25 = 0.032 M, option (b).

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