If `N_(2)` gas is bubbled through water at `293 K`, how many millimoles of `N_(2)` gas would dissolve in`1 L` of water. Assume that` N_(2)` exerts a partial pressure of 0.987 bar. Given that Henry law constant for `N_(2)` at `293 K` is 76.48 kbar.

Explanation

To solve the problem of how many millimoles of \( N_2 \) gas would dissolve in 1 L of water at 293 K, given a partial pressure of 0.987 bar and a Henry's law constant of 76.48 kbar, we can follow these steps:

### Step 1: Convert Henry's Law Constant to Bar
The given Henry's law constant is in kilobars, so we need to convert it to bars for consistency with the pressure units.

\[ K_H = 76.48 \, \text{kbar} = 76.48 \times 1000 \, \text{bar} = 76480 \, \text{bar} \]

**Hint:** Remember that 1 kbar = 1000 bar.

### Step 2: Apply Henry's Law
According to Henry's law, the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula is:

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

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

Substituting the values:

\[ C = \frac{0.987 \, \text{bar}}{76480 \, \text{bar}} = 1.29 \times 10^{-5} \, \text{mol/L} \]

**Hint:** Make sure to use the correct units for pressure and the Henry's law constant.

### Step 3: Calculate Moles of \( N_2 \)
Since we are considering 1 L of water, the number of moles of \( N_2 \) that dissolve is equal to the concentration calculated:

\[ \text{Moles of } N_2 = C \times \text{Volume} = 1.29 \times 10^{-5} \, \text{mol/L} \times 1 \, \text{L} = 1.29 \times 10^{-5} \, \text{mol} \]

**Hint:** Remember that the concentration is in moles per liter, so multiplying by the volume in liters gives you moles.

### Step 4: Convert Moles to Millimoles
To convert moles to millimoles, multiply by 1000:

\[ \text{Millimoles of } N_2 = 1.29 \times 10^{-5} \, \text{mol} \times 1000 = 0.0129 \, \text{mmol} \]

**Hint:** To convert from moles to millimoles, remember that 1 mole = 1000 millimoles.

### Final Answer
The number of millimoles of \( N_2 \) gas that would dissolve in 1 L of water at 293 K is approximately:

\[ \text{Millimoles of } N_2 \approx 0.0129 \, \text{mmol} \]