1. Write the balanced redox reaction

In acidic medium, potassium permanganate oxidizes oxalic acid:

$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$

So, the mole ratio is:

$2\text{ mol } KMnO_4 : 5\text{ mol oxalic acid}$

2. Moles of $KMnO_4$ used

Given:

• Volume of $KMnO_4 = 10.0\,\text{mL} = 0.0100\,\text{L}$
• Molarity of $KMnO_4 = 0.05\,M$

Therefore,

$\text{moles of } KMnO_4 = M \times V = 0.05 \times 0.0100 = 5.0 \times 10^{-4}$

3. Moles of oxalic acid dihydrate reacted

Using the stoichiometric ratio:

$\frac{\text{moles of oxalic acid}}{\text{moles of } KMnO_4} = \frac{5}{2}$

Hence,

$\text{moles of oxalic acid dihydrate} = \frac{5}{2} \times 5.0 \times 10^{-4} = 1.25 \times 10^{-3}$

These moles are present in $10.0\,\text{mL}$ of the given solution.

4. Find molarity of oxalic acid dihydrate solution

Volume of oxalic acid solution $= 10.0\,\text{mL} = 0.0100\,\text{L}$

$M = \frac{1.25 \times 10^{-3}}{0.0100} = 0.125\,M$

5. Calculate strength in g/L

Given oxalic acid is oxalic acid dihydrate:

$H_2C_2O_4\cdot 2H_2O$

Molar mass:

$= 2(1) + 2(12) + 4(16) + 2(18) = 2 + 24 + 64 + 36 = 126\,\text{g/mol}$

Strength:

$\text{Strength} = M \times \text{molar mass} = 0.125 \times 126 = 15.75\,\text{g/L}$

6. Match with the required format

They ask for:

$\underline{\hspace{2cm}} \times 10^{-2}\,\text{g/L}$

Now,

$15.75\,\text{g/L} = 1575 \times 10^{-2}\,\text{g/L}$

Rounded to the nearest integer, the required number is:

$\boxed{1575}$

7. Comparison with stored answer

Stored correct answer = $1575$

My derived answer also = $1575$

So the answer agrees.