Excess of Kl and dil. `H_2SO_4` were mixed in 50 mL `H_2O_2`. The liberated `l_2` required 20 mL 0.1 N `Na_2S_2O_3`. Find out the strength of `H_2O_2` in g `"litre"^(-1)`.

Explanation

To find the strength of \( H_2O_2 \) in g/L, we can follow these steps:

### Step 1: Determine the equivalent weight of \( H_2O_2 \)
The molecular weight of \( H_2O_2 \) is calculated as follows:
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Therefore, the molecular weight of \( H_2O_2 \) = 2 + 32 = 34 g/mol.

The equivalent weight of \( H_2O_2 \) is given by:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
where \( n \) is the number of electrons transferred in the reaction. For \( H_2O_2 \), \( n = 2 \) (as it can release 2 moles of electrons). Thus, the equivalent weight is:
\[ \text{Equivalent weight of } H_2O_2 = \frac{34}{2} = 17 \text{ g/equiv} \]

### Step 2: Calculate the equivalents of \( I_2 \) liberated
The volume of \( Na_2S_2O_3 \) used is 20 mL with a normality of 0.1 N. The number of equivalents of \( Na_2S_2O_3 \) can be calculated as:
\[ \text{Equivalents of } Na_2S_2O_3 = \text{Volume (L)} \times \text{Normality} = 0.020 \, \text{L} \times 0.1 \, \text{N} = 0.002 \, \text{equivalents} \]

### Step 3: Relate the equivalents of \( I_2 \) to \( H_2O_2 \)
From the reaction, we know that 1 equivalent of \( H_2O_2 \) produces 1 equivalent of \( I_2 \). Therefore, the equivalents of \( H_2O_2 \) that reacted to liberate \( I_2 \) is also 0.002 equivalents.

### Step 4: Calculate the mass of \( H_2O_2 \) in the solution
Using the equivalent weight calculated earlier, we can find the mass of \( H_2O_2 \):
\[ \text{Mass of } H_2O_2 = \text{Equivalents} \times \text{Equivalent weight} = 0.002 \, \text{equivalents} \times 17 \, \text{g/equiv} = 0.034 \, \text{g} \]

### Step 5: Calculate the strength of \( H_2O_2 \) in g/L
Since the mass of \( H_2O_2 \) is found in 50 mL of solution, we need to convert this to a per liter basis:
\[ \text{Strength of } H_2O_2 = \frac{0.034 \, \text{g}}{50 \, \text{mL}} \times 1000 \, \text{mL/L} = 0.68 \, \text{g/L} \]

### Final Answer
The strength of \( H_2O_2 \) is **0.68 g/L**.

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