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, we will use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of \( H_2S \) is calculated as follows:
- Hydrogen (H) = 1 g/mol, and there are 2 H atoms: \( 2 \times 1 = 2 \) g/mol
- Sulfur (S) = 32 g/mol

Thus, the molar mass of \( H_2S \) is:
\[ 2 + 32 = 34 \text{ g/mol} \]
Now, substituting the values:
\[ \text{Number of moles of } H_2S = \frac{0.81 \text{ g}}{34 \text{ g/mol}} \approx 0.0238 \text{ moles} \]

### Step 2: Determine the n-factor for \( H_2S \)
The n-factor is determined by the change in oxidation state of sulfur in \( H_2S \). In \( H_2S \), sulfur has an oxidation state of -2, and in elemental sulfur (S), it has an oxidation state of 0. The change in oxidation state is:
\[ \text{Change} = 0 - (-2) = 2 \]
Thus, the n-factor for \( H_2S \) is 2.

### Step 3: Calculate the equivalence of \( H_2S \)
Using the formula:
\[ \text{Equivalence} = \text{n-factor} \times \text{number of moles} \]
Substituting the values:
\[ \text{Equivalence of } H_2S = 2 \times 0.0238 \approx 0.0476 \text{ equivalents} \]

### Step 4: Set up the equivalence equation
According to the stoichiometry of the reaction, the equivalences of \( K_2Cr_2O_7 \) will equal the equivalences of \( H_2S \):
\[ \text{Equivalence of } K_2Cr_2O_7 = \text{Equivalence of } H_2S \]
The equivalence of \( K_2Cr_2O_7 \) can be expressed as:
\[ \text{Equivalence} = \text{Normality} \times \text{Volume (L)} \]
Given that the normality of \( K_2Cr_2O_7 \) is 2 N, we have:
\[ 2 \times V = 0.0476 \]

### Step 5: Solve for the volume \( V \)
Rearranging the equation gives:
\[ V = \frac{0.0476}{2} = 0.0238 \text{ L} \]
To convert this to milliliters:
\[ V = 0.0238 \text{ L} \times 1000 \text{ mL/L} = 23.8 \text{ mL} \]

### Final Answer
The volume of 2 N \( K_2Cr_2O_7 \) solution required to oxidize 0.81 g of \( H_2S \) is **23.8 mL**.

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