Cacluate the amount of `CO_(2)` dissolved at 4 atm in 1 `"dm"^(3)` of water at 298 K . The Henry's law constant for `CO_(2)` at 298K is 1.67 k bar .

Explanation

To solve the problem of calculating the amount of \( CO_2 \) dissolved at 4 atm in 1 dm³ of water at 298 K, we will use Henry's Law and follow these steps:

### Step 1: Understand Henry's Law
Henry's Law states that the amount of gas dissolved in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid. The formula is given by:

\[ P = k_H \cdot X \]

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

### Step 2: Convert Henry's Law Constant
Given \( k_H = 1.67 \, \text{kbar} \), we need to convert it to atm for consistency.

\[ 1 \, \text{kbar} = 1000 \, \text{bar} = 1000 \times 0.98692 \, \text{atm} \approx 986.92 \, \text{atm} \]

Thus,

\[ k_H = 1.67 \, \text{kbar} = 1.67 \times 1000 \, \text{bar} = 1670 \, \text{bar} \approx 1670 \times 0.98692 \, \text{atm} \approx 1648.57 \, \text{atm} \]

### Step 3: Calculate the Mole Fraction
Given the partial pressure \( P = 4 \, \text{atm} \):

Using Henry's Law:

\[ 4 \, \text{atm} = 1648.57 \, \text{atm} \cdot X \]

Rearranging to find \( X \):

\[ X = \frac{4 \, \text{atm}}{1648.57 \, \text{atm}} \approx 0.00243 \]

### Step 4: Calculate Moles of Water
The volume of water is 1 dm³ (or 1 L). The density of water is approximately 1 g/cm³, so:

\[ \text{Mass of water} = 1000 \, \text{g} \]

The molar mass of water \( H_2O \) is approximately 18 g/mol. Therefore, the number of moles of water \( n_{H_2O} \) is:

\[ n_{H_2O} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 5: Calculate Moles of \( CO_2 \)
Using the mole fraction formula:

\[ X = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \]

Substituting the values we have:

\[ 0.00243 = \frac{n_{CO_2}}{n_{CO_2} + 55.56} \]

Let \( n_{CO_2} = x \):

\[ 0.00243 = \frac{x}{x + 55.56} \]

Cross-multiplying gives:

\[ 0.00243(x + 55.56) = x \]

Expanding and rearranging:

\[ 0.00243x + 0.135 = x \]

\[ 0.135 = x - 0.00243x \]

\[ 0.135 = 0.99757x \]

\[ x \approx \frac{0.135}{0.99757} \approx 0.135 \]

### Step 6: Calculate Mass of \( CO_2 \) Dissolved
The molar mass of \( CO_2 \) is approximately 44 g/mol. Therefore, the mass of \( CO_2 \) dissolved is:

\[ \text{Mass} = n_{CO_2} \times \text{Molar mass of } CO_2 = 0.135 \, \text{mol} \times 44 \, \text{g/mol} \approx 5.94 \, \text{g} \]

### Final Answer
The amount of \( CO_2 \) dissolved at 4 atm in 1 dm³ of water at 298 K is approximately **5.94 g**.

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