`H_(2)S`, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of `H_(2)S` in water at `STP` is `0.195 m`, calculate Henry's law constant.

Explanation

To calculate Henry's law constant for \( H_2S \), we can follow these steps:

### Step 1: Understand the given data
We are given the solubility of \( H_2S \) in water at STP, which is \( 0.195 \, m \) (molality). This means that \( 0.195 \) moles of \( H_2S \) are dissolved in \( 1 \, kg \) of water.

### Step 2: Calculate the moles of water
To find the moles of water, we use the molar mass of water, which is \( 18 \, g/mol \).

\[ \text{Moles of water} = \frac{\text{mass of water (g)}}{\text{molar mass of water (g/mol)}} \]

\[ \text{Moles of water} = \frac{1000 \, g}{18 \, g/mol} \approx 55.55 \, moles \]

### Step 3: Calculate the mole fraction of \( H_2S \)
The mole fraction (\( X \)) of \( H_2S \) can be calculated using the formula:

\[ X_{H_2S} = \frac{\text{moles of } H_2S}{\text{moles of } H_2S + \text{moles of water}} \]

Substituting the values:

\[ X_{H_2S} = \frac{0.195}{0.195 + 55.55} \approx \frac{0.195}{55.745} \approx 0.0034 \]

### Step 4: Use Henry's Law to find the constant
According to Henry's Law, the partial pressure of the gas above the solution is directly proportional to its mole fraction in the solution:

\[ P = k_H \cdot X \]

Where:
- \( P \) is the partial pressure of the gas (at STP, \( P = 1 \, atm = 0.987 \, bar \))
- \( k_H \) is Henry's law constant
- \( X \) is the mole fraction of the gas

Rearranging the equation to solve for \( k_H \):

\[ k_H = \frac{P}{X} \]

Substituting the values:

\[ k_H = \frac{0.987 \, bar}{0.0034} \approx 290.29 \, bar \]

### Final Answer
The Henry's law constant for \( H_2S \) is approximately \( 290.29 \, bar \).

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