10 mL solution of `H_(2)SO_(4)` and `H_(2)C_(2)O_(4)` (oxalic acid), on titration with with 0.1 M KOH, required 20 " mL of " the base.10 " mL of " the same solution on titration with `(M)/(300)K_(2)Cr_(2)O_(7)` required 50 " mL of " `K_(2)Cr_(2)O_(7)`.
Q. Strength of oxalic acid in the solution is:

Options

A. `4.5gL^(-1)`
B. `4.9gL^(-1)`
C. `2.25gL^(-1)`
D. `2.45gL^(-1)`

Explanation

To solve the problem, we need to determine the strength of oxalic acid (H₂C₂O₄) in a solution that also contains sulfuric acid (H₂SO₄). We'll use the information provided from the titrations with KOH and K₂Cr₂O₇ to find the required strength.

### Step-by-Step Solution:

1. **Identify the Reactions and N-Factors:**
- For **H₂SO₄**: It dissociates to give 2 H⁺ ions. Therefore, its n-factor is **2**.
- For **H₂C₂O₄**: It dissociates to give 2 H⁺ ions as well. Thus, its n-factor is also **2**.
- For **KOH**: It provides 1 OH⁻ ion, so its n-factor is **1**.
- For **K₂Cr₂O₇**: In acidic medium, it undergoes reduction and has an n-factor of **6**.

2. **Calculate Milliequivalents of KOH:**
- The total volume of KOH used is **20 mL** with a concentration of **0.1 M**.
- Milliequivalents of KOH = Volume (mL) × Molarity × n-factor
\[ \text{Milliequivalents of KOH} = 20 \, \text{mL} \times 0.1 \, \text{mol/L} \times 1 = 2 \, \text{meq} \]

3. **Set Up the Equation for the First Titration:**
- Let **X** be the milliequivalents of H₂C₂O₄ and **Y** be the milliequivalents of H₂SO₄.
- From the first titration with KOH:
\[ X + Y = 2 \quad \text{(Equation 1)} \]

4. **Calculate Milliequivalents of K₂Cr₂O₇:**
- The total volume of K₂Cr₂O₇ used is **50 mL** with a concentration of **1/300 M**.
- Milliequivalents of K₂Cr₂O₇ = Volume (mL) × Molarity × n-factor
\[ \text{Milliequivalents of K₂Cr₂O₇} = 50 \, \text{mL} \times \frac{1}{300} \, \text{mol/L} \times 6 = 1 \, \text{meq} \]

5. **Set Up the Equation for the Second Titration:**
- Since only oxalic acid reacts with K₂Cr₂O₇:
\[ X = 1 \quad \text{(Equation 2)} \]

6. **Solve the Equations:**
- Substitute Equation 2 into Equation 1:
\[ 1 + Y = 2 \implies Y = 1 \]

7. **Determine the Strength of Oxalic Acid:**
- The milliequivalents of H₂C₂O₄ is **1** for **10 mL** of the solution.
- To find the weight of H₂C₂O₄:
\[ \text{Weight of H₂C₂O₄} = \text{milliequivalents} \times \text{molecular weight} \div \text{n-factor} \]
- The molecular weight of H₂C₂O₄ is **90 g/mol**.
\[ \text{Weight} = 1 \times 90 \, \text{g/mol} \div 2 = 45 \, \text{g} \]

8. **Calculate the Strength:**
- Strength (g/L) = Weight (g) / Volume (L)
\[ \text{Strength of H₂C₂O₄} = \frac{45 \, \text{g}}{0.01 \, \text{L}} = 4500 \, \text{g/L} \]
- Since we need the strength in grams per liter for the original 10 mL solution:
\[ \text{Strength} = \frac{0.045 \, \text{g}}{0.01 \, \text{L}} = 4.5 \, \text{g/L} \]

### Final Answer:
The strength of oxalic acid in the solution is **4.5 g/L**.