When a gas is bubbled through water at 298 K, a very dilute solution of gas is obtained. Henry's law constant for the gas is 100 kbar. If gas exerts a pressure of 1 bar, the number of moles of gas dissolved in 1 litre of water is `x xx 10^(-5)`. Find the value of x here?
Report your answer upto two decimal places.

Answer

To solve the problem, we will use Henry's law, which states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid. The formula for Henry's law can be expressed as:

\[ P = K_H \cdot X_A \]

Where:
- \( P \) = partial pressure of the gas
- \( K_H \) = Henry's law constant
- \( X_A \) = mole fraction of the gas in the solution

### Step-by-Step Solution:

1. **Identify Given Values:**
- Henry's law constant \( K_H = 100 \, \text{kbar} = 100 \times 10^5 \, \text{Pa} \) (since 1 kbar = \( 10^5 \) Pa)
- Partial pressure of the gas \( P = 1 \, \text{bar} = 10^5 \, \text{Pa} \)

2. **Convert Henry's Law Constant to Bar:**
- Since \( K_H \) is given in kbar, we convert it to bar:
\[ K_H = 100 \, \text{kbar} = 100 \times 10^5 \, \text{Pa} = 100 \, \text{bar} \]

3. **Set Up the Equation Using Henry's Law:**
\[ P = K_H \cdot X_A \]
Substituting the values we have:
\[ 1 \, \text{bar} = 100 \, \text{bar} \cdot X_A \]

4. **Solve for Mole Fraction \( X_A \):**
\[ X_A = \frac{1 \, \text{bar}}{100 \, \text{bar}} = 0.01 \]

5. **Calculate Moles of Water in 1 Liter:**
- The molarity of water is approximately \( 55.55 \, \text{mol/L} \).

6. **Relate Mole Fraction to Moles:**
- The mole fraction \( X_A \) can also be expressed in terms of moles of gas \( n_g \) and moles of water \( n_w \):
\[ X_A = \frac{n_g}{n_g + n_w} \]
Since \( n_w \) is much larger than \( n_g \) (very dilute solution), we can approximate:
\[ X_A \approx \frac{n_g}{n_w} \]

7. **Substituting Values:**
- Let \( n_w = 55.55 \, \text{mol} \) (moles of water in 1 liter):
\[ 0.01 \approx \frac{n_g}{55.55} \]

8. **Solve for Moles of Gas \( n_g \):**
\[ n_g = 0.01 \times 55.55 = 0.5555 \, \text{mol} \]

9. **Express in Scientific Notation:**
- The number of moles of gas dissolved in 1 liter of water is:
\[ n_g = 0.5555 \, \text{mol} = 55.55 \times 10^{-5} \, \text{mol} \]

10. **Identify the Value of \( x \):**
- From the expression \( n_g = x \times 10^{-5} \), we can see that:
\[ x = 55.55 \]

### Final Answer:
The value of \( x \) is \( 55.55 \).