What volume of 2 N ` K_(2)Cr_(2)O_(7)` solution is required to oxidise 0.81 g of `H_(2)S` in acidic medium. ` K_(2)Cr_(2)O_(7) + H_(2)S + H_(2)SO_(4) rarr S + K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) `

Explanation

To solve the problem of determining the volume of 2 N \( K_2Cr_2O_7 \) solution required to oxidize 0.81 g of \( H_2S \) in acidic medium, we will follow these steps:

### Step 1: Calculate the number of moles of \( H_2S \)

To find the number of moles of \( H_2S \), we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
The molar mass of \( H_2S \) is calculated as follows:
- Hydrogen (H) = 1 g/mol (2 atoms) = 2 g/mol
- Sulfur (S) = 32 g/mol

Thus, the molar mass of \( H_2S \) is:
\[ \text{Molar mass of } H_2S = 2 + 32 = 34 \text{ g/mol} \]

Now, substituting the values:
\[ \text{Number of moles of } H_2S = \frac{0.81 \text{ g}}{34 \text{ g/mol}} \approx 0.02382 \text{ moles} \]

### Step 2: Determine the n-factor for \( H_2S \)

The n-factor is determined by the change in oxidation state of sulfur in \( H_2S \) to elemental sulfur (S).

- In \( H_2S \), the oxidation state of sulfur is -2.
- In elemental sulfur (S), the oxidation state is 0.

Thus, the change in oxidation state is:
\[ \text{Change} = 0 - (-2) = 2 \]
Therefore, the n-factor for \( H_2S \) is 2.

### Step 3: Calculate the equivalence of \( H_2S \)

Using the n-factor, we can calculate the equivalence of \( H_2S \):
\[ \text{Equivalence of } H_2S = \text{n-factor} \times \text{number of moles} \]
\[ \text{Equivalence of } H_2S = 2 \times 0.02382 \approx 0.04764 \]

### Step 4: Relate the equivalence of \( K_2Cr_2O_7 \) to \( H_2S \)

From the stoichiometry of the reaction, we know that the equivalence of \( K_2Cr_2O_7 \) is equal to the equivalence of \( H_2S \):
\[ \text{Equivalence of } K_2Cr_2O_7 = \text{Equivalence of } H_2S \]

### Step 5: Calculate the volume of \( K_2Cr_2O_7 \) solution required

The equivalence of \( K_2Cr_2O_7 \) can also be expressed in terms of its normality (N) and volume (V):
\[ \text{Equivalence} = N \times V \]
Given that the normality of \( K_2Cr_2O_7 \) is 2 N, we can set up the equation:
\[ 2 \times V = 0.04764 \]

Solving for V:
\[ V = \frac{0.04764}{2} \approx 0.02382 \text{ L} \]
Converting liters to milliliters:
\[ V \approx 0.02382 \times 1000 \approx 23.82 \text{ mL} \]

### Final Answer:
The volume of 2 N \( K_2Cr_2O_7 \) solution required to oxidize 0.81 g of \( H_2S \) in acidic medium is approximately **23.82 mL**.

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