10.0 mL of 0.05 M KMnO4 solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ........ `xx 10 ^(-2) g // L.`
(Round off to the nearest integer)

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of KMnO4 used in the titration.
Given:
- Volume of KMnO4 solution = 10.0 mL = 0.010 L
- Molarity of KMnO4 solution = 0.05 M

Using the formula:
\[ \text{Moles of KMnO4} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Moles of KMnO4} = 0.05 \, \text{mol/L} \times 0.010 \, \text{L} = 0.0005 \, \text{mol} \]

### Step 2: Write the balanced chemical equation for the reaction.
The reaction between KMnO4 and oxalic acid (H2C2O4·2H2O) in acidic medium is:
\[ \text{KMnO4} + \text{H2C2O4} \rightarrow \text{Mn}^{2+} + \text{CO2} + \text{H2O} \]
From the balanced equation, we see that 1 mole of KMnO4 reacts with 5 moles of oxalic acid.

### Step 3: Calculate the moles of oxalic acid that reacted.
Using the stoichiometry from the balanced equation:
\[ \text{Moles of oxalic acid} = 5 \times \text{Moles of KMnO4} \]
\[ \text{Moles of oxalic acid} = 5 \times 0.0005 \, \text{mol} = 0.0025 \, \text{mol} \]

### Step 4: Calculate the concentration of oxalic acid in the solution.
The volume of the oxalic acid solution used is also 10.0 mL = 0.010 L. Therefore, the molarity of the oxalic acid solution is:
\[ \text{Molarity of oxalic acid} = \frac{\text{Moles of oxalic acid}}{\text{Volume (in L)}} \]
\[ \text{Molarity of oxalic acid} = \frac{0.0025 \, \text{mol}}{0.010 \, \text{L}} = 0.25 \, \text{M} \]

### Step 5: Calculate the strength of the oxalic acid solution in g/L.
The molar mass of oxalic acid dihydrate (H2C2O4·2H2O) is approximately:
- C: 12.01 g/mol × 2 = 24.02 g/mol
- H: 1.008 g/mol × 2 = 2.016 g/mol
- O: 16.00 g/mol × 4 = 64.00 g/mol
- Water: 18.02 g/mol × 2 = 36.04 g/mol

Total molar mass = 24.02 + 2.016 + 64.00 + 36.04 = 126.08 g/mol

Now, we can calculate the strength in g/L:
\[ \text{Strength (g/L)} = \text{Molarity} \times \text{Molar mass} \]
\[ \text{Strength (g/L)} = 0.25 \, \text{mol/L} \times 126.08 \, \text{g/mol} = 31.52 \, \text{g/L} \]

### Step 6: Round off to the nearest integer.
The strength of the given oxalic acid solution is approximately 32 g/L.

### Final Answer:
The strength of the given oxalic acid solution is **32 × 10^(-2) g/L**.

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