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

Options

A. 0.555
B. `55.55xx10^(-5)`
C. `55.55xx10^(-3)`
D. `5.55xx10^(-5)`

Explanation

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

\[ C = \frac{P}{K_H} \]

Where:
- \( C \) is the concentration of the gas in the solution (in moles per liter),
- \( P \) is the pressure of the gas (in bar),
- \( K_H \) is Henry's Law constant (in bar).

### Step 1: Identify the given values
- Pressure of the gas, \( P = 1 \, \text{bar} \)
- Henry's Law constant, \( K_H = 100 \, \text{kbar} = 100 \times 10^3 \, \text{bar} = 100000 \, \text{bar} \)

### Step 2: Apply Henry's Law
Using the formula from Henry's Law:

\[ C = \frac{P}{K_H} \]

Substituting the values:

\[ C = \frac{1 \, \text{bar}}{100000 \, \text{bar}} = \frac{1}{100000} \, \text{mol/L} \]

### Step 3: Calculate the concentration
Calculating the concentration:

\[ C = 1 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Determine the number of moles in 1 liter of water
Since the concentration \( C \) is in moles per liter, the number of moles of gas dissolved in 1 liter of water is simply:

\[ \text{Number of moles of gas} = C \times \text{Volume of solution} \]

Given that the volume of the solution is 1 liter:

\[ \text{Number of moles of gas} = 1 \times 10^{-5} \, \text{mol/L} \times 1 \, \text{L} = 1 \times 10^{-5} \, \text{mol} \]

### Final Answer
The number of moles of gas dissolved in 1 liter of water is:

\[ \text{Number of moles of gas} = 1 \times 10^{-5} \, \text{mol} \]

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