henry's law constant for `CO_(2)` in water is `2.5xx10^(8)` Pa at `298K` calcualte mmole of `CO_(2)` dissolbed in 144 g water at 2.5 atm pressure at 298K. [take 1 atm `=10^(6)(N)/(m^(2))` or Pa]

Answer

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 is given by:

\[ P_{CO_2} = k_H \cdot X_{CO_2} \]

Where:
- \( P_{CO_2} \) is the partial pressure of \( CO_2 \) (in Pascals),
- \( k_H \) is Henry's law constant (in Pascals),
- \( X_{CO_2} \) is the mole fraction of \( CO_2 \).

### Step 1: Convert the pressure from atm to Pa
Given:
- Pressure \( P_{CO_2} = 2.5 \) atm

Using the conversion \( 1 \text{ atm} = 10^6 \text{ Pa} \):

\[ P_{CO_2} = 2.5 \times 10^6 \text{ Pa} = 2.5 \times 10^5 \text{ Pa} \]

### Step 2: Calculate the mole fraction of \( CO_2 \)
Using Henry's law:

\[ X_{CO_2} = \frac{P_{CO_2}}{k_H} \]

Substituting the values:

\[ X_{CO_2} = \frac{2.5 \times 10^5 \text{ Pa}}{2.5 \times 10^8 \text{ Pa}} = 10^{-3} \]

### Step 3: Calculate moles of water
Given:
- Mass of water = 144 g
- Molar mass of water \( H_2O \) = 18 g/mol

Calculating moles of water:

\[ n_{H_2O} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{144 \text{ g}}{18 \text{ g/mol}} = 8 \text{ moles} \]

### Step 4: Calculate moles of \( CO_2 \)
Using the mole fraction calculated in Step 2:

\[ X_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}} \quad \text{(since } n_{CO_2} \text{ is very small)} \]

Rearranging gives:

\[ n_{CO_2} = X_{CO_2} \cdot n_{H_2O} \]

Substituting the values:

\[ n_{CO_2} = 10^{-3} \cdot 8 = 8 \times 10^{-3} \text{ moles} \]

### Step 5: Convert moles of \( CO_2 \) to millimoles
Since 1 mole = 1000 millimoles:

\[ n_{CO_2} = 8 \times 10^{-3} \text{ moles} = 8 \text{ millimoles} \]

### Final Answer:
The amount of \( CO_2 \) dissolved in 144 g of water at 2.5 atm pressure at 298 K is **8 millimoles**.

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