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. The strength of `H_(2)SO_(4)` is the solution is:

Options

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

Explanation

To find the strength of \( H_2SO_4 \) in the solution, we can follow these steps:

### Step 1: Calculate the milliequivalents of \( H_2SO_4 \) using KOH titration

The balanced reaction for the titration of \( H_2SO_4 \) with KOH is:

\[ H_2SO_4 + 2 KOH \rightarrow K_2SO_4 + 2 H_2O \]

From the reaction, we see that 1 mole of \( H_2SO_4 \) reacts with 2 moles of KOH.

Given:
- Volume of KOH used = 20 mL = 0.020 L
- Molarity of KOH = 0.1 M

Calculate the moles of KOH used:

\[ \text{Moles of KOH} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} \]

Since 2 moles of KOH react with 1 mole of \( H_2SO_4 \), the moles of \( H_2SO_4 \) can be calculated as:

\[ \text{Moles of } H_2SO_4 = \frac{0.002 \, \text{mol}}{2} = 0.001 \, \text{mol} \]

### Step 2: Calculate the milliequivalents of \( H_2SO_4 \)

Milliequivalents of \( H_2SO_4 \):

\[ \text{Milliequivalents} = \text{Moles} \times \text{n} \times 1000 = 0.001 \, \text{mol} \times 2 \times 1000 = 2 \, \text{meq} \]

### Step 3: Calculate the weight of \( H_2SO_4 \)

The equivalent weight of \( H_2SO_4 \) is calculated as follows:

\[ \text{Equivalent weight of } H_2SO_4 = \frac{\text{Molecular weight}}{n} = \frac{98 \, \text{g/mol}}{2} = 49 \, \text{g/equiv} \]

Now, using the milliequivalents calculated:

\[ \text{Weight of } H_2SO_4 = \text{Milliequivalents} \times \text{Equivalent weight} \times 10^{-3} = 2 \, \text{meq} \times 49 \, \text{g/equiv} \times 10^{-3} = 0.098 \, \text{g} \]

### Step 4: Calculate the strength of \( H_2SO_4 \) in g/L

Since the total volume of the solution is 10 mL (0.010 L), we can convert the weight to grams per liter:

\[ \text{Strength of } H_2SO_4 = \frac{0.098 \, \text{g}}{0.010 \, \text{L}} = 9.8 \, \text{g/L} \]

### Step 5: Final Calculation

To express the strength in g/L, we multiply by 1000 to convert from mL to L:

\[ \text{Strength of } H_2SO_4 = 9.8 \, \text{g/L} \]

### Conclusion

The strength of \( H_2SO_4 \) in the solution is **9.8 g/L**.

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