A sample which conains exactly 0.5 g of uranium in the form of `U^(4+)` . The total uranium is allowed to oxidized by 50 ml of `KMnO_(4)` . The reaction taking place is
`U^(4+) + KMnO_(4) + H_(2)O to UO_(2)^(2+) + Mn^(2+) +H_(2)O^(+)`
Find the concentration of `KMnO_(4)` required for the above purpose [U=238]

Options

A. 0.0336 M
B. 0.0084 M
C. 0.0168 M
D. 0.0672 M

Answer

To find the concentration of KMnO4 required to oxidize 0.5 g of uranium in the form of U^(4+), we will follow these steps:

### Step 1: Calculate the number of moles of uranium (U^(4+))
Given:
- Mass of uranium = 0.5 g
- Atomic mass of uranium (U) = 238 g/mol

Using the formula for moles:
\[ \text{Number of moles of U} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.5 \, \text{g}}{238 \, \text{g/mol}} = 0.0021 \, \text{mol} \]

### Step 2: Determine the n-factor for U^(4+)
In the reaction:
\[ U^{4+} \rightarrow UO_2^{2+} \]
The oxidation state of uranium changes from +4 to +6. The change in oxidation state is:
\[ \Delta \text{oxidation state} = 6 - 4 = 2 \]
Thus, the n-factor (valency factor) for U^(4+) is 2.

### Step 3: Calculate the number of equivalents of uranium
\[ \text{Number of equivalents of U} = \text{Number of moles} \times n \text{-factor} = 0.0021 \, \text{mol} \times 2 = 0.0042 \, \text{equivalents} \]

### Step 4: Determine the n-factor for KMnO4
In KMnO4, manganese (Mn) is in the +7 oxidation state and is reduced to +2:
\[ \Delta \text{oxidation state} = 7 - 2 = 5 \]
Thus, the n-factor for KMnO4 is 5.

### Step 5: Calculate the number of equivalents of KMnO4 required
Since the number of equivalents of U^(4+) is equal to the number of equivalents of KMnO4:
\[ \text{Number of equivalents of KMnO4} = 0.0042 \, \text{equivalents} \]

### Step 6: Calculate the number of moles of KMnO4
Using the n-factor of KMnO4:
\[ \text{Number of moles of KMnO4} = \frac{\text{Number of equivalents}}{n \text{-factor}} = \frac{0.0042 \, \text{equivalents}}{5} = 0.00084 \, \text{mol} \]

### Step 7: Calculate the concentration of KMnO4
Given that the volume of KMnO4 solution used is 50 mL (or 0.050 L):
\[ \text{Concentration (C)} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.00084 \, \text{mol}}{0.050 \, \text{L}} = 0.0168 \, \text{mol/L} \]

### Final Answer
The concentration of KMnO4 required is **0.0168 M**.

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