The Henry's law constant for `CO_(2)` in water at 298 K is 1.67 kbar. Calculate the solubility of `CO_(2)` at 298 K when the pressure of `CO_(2)` is one bar.

Explanation

To solve the problem, we will use Henry's law, which relates the solubility of a gas in a liquid to the partial pressure of that gas above the liquid. The formula for Henry's law is given by:

\[ P = k_H \cdot X \]

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

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Henry's law constant for \( CO_2 \) in water at 298 K: \( k_H = 1.67 \, \text{kbar} = 1670 \, \text{bar} \)
- Pressure of \( CO_2 \): \( P = 1 \, \text{bar} \)

**Step 2: Rearrange Henry's law to find the mole fraction \( X \).**
Using the formula:
\[ X = \frac{P}{k_H} \]

Substituting the known values:
\[ X = \frac{1 \, \text{bar}}{1670 \, \text{bar}} \]
\[ X = \frac{1}{1670} \]

**Step 3: Calculate the mole fraction \( X \).**
\[ X \approx 0.0005988 \]

**Step 4: Calculate the mass of water in 1 liter (1000 mL).**
Since the density of water is approximately \( 1 \, \text{g/mL} \):
\[ \text{Mass of water} = 1000 \, \text{g} \]

**Step 5: Calculate the number of moles of water.**
Using the molar mass of water (\( H_2O \)) which is \( 18 \, \text{g/mol} \):
\[ \text{Moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

**Step 6: Use the mole fraction formula to find the moles of \( CO_2 \).**
The mole fraction is defined as:
\[ X = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \]

Where \( n_{CO_2} \) is the number of moles of \( CO_2 \) and \( n_{H_2O} \) is the number of moles of water.

Substituting the values we have:
\[ \frac{1}{1670} = \frac{n_{CO_2}}{n_{CO_2} + 55.56} \]

**Step 7: Solve for \( n_{CO_2} \).**
Cross-multiplying gives:
\[ n_{CO_2} = \frac{1}{1670} (n_{CO_2} + 55.56) \]

Rearranging:
\[ 1670 n_{CO_2} = n_{CO_2} + 55.56 \]
\[ 1669 n_{CO_2} = 55.56 \]
\[ n_{CO_2} = \frac{55.56}{1669} \approx 0.0333 \, \text{mol} \]

**Step 8: Calculate the solubility of \( CO_2 \) in mol/L.**
Since we are considering 1 liter of water, the solubility of \( CO_2 \) is:
\[ \text{Solubility} = n_{CO_2} \approx 0.0333 \, \text{mol/L} \]

### Final Answer:
The solubility of \( CO_2 \) at 298 K when the pressure of \( CO_2 \) is 1 bar is approximately \( 0.0333 \, \text{mol/L} \).

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