1 litre of water under a nitrogen pressure of 1 bar dissolves `2xx10^(-5)` kg of nitrogen at 293 K. Calculate Henry's law constant.

Explanation

To calculate Henry's law constant (K_H) for the dissolution of nitrogen in water, we can follow these steps:

### Step 1: Determine the mass of water
Given that we have 1 liter of water, we can convert this volume to mass. The density of water is approximately 1 g/mL, so:
\[ \text{Mass of water} = 1 \, \text{liter} = 1000 \, \text{g} \]

### Step 2: Calculate the moles of water
To find the moles of water, we use the molar mass of water (H₂O), which is approximately 18 g/mol:
\[ \text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 3: Determine the mass of nitrogen
The mass of nitrogen dissolved is given as \(2 \times 10^{-5} \, \text{kg}\). To convert this to grams:
\[ \text{Mass of nitrogen} = 2 \times 10^{-5} \, \text{kg} = 2 \times 10^{-2} \, \text{g} \]

### Step 4: Calculate the moles of nitrogen
Using the molar mass of nitrogen (N₂), which is approximately 28 g/mol:
\[ \text{Moles of nitrogen} = \frac{\text{Mass of nitrogen}}{\text{Molar mass of nitrogen}} = \frac{2 \times 10^{-2} \, \text{g}}{28 \, \text{g/mol}} \approx 7.14 \times 10^{-4} \, \text{mol} \]

### Step 5: Calculate the mole fraction of nitrogen
The mole fraction (X) of nitrogen in the solution can be calculated using the formula:
\[ X_{N_2} = \frac{\text{Moles of nitrogen}}{\text{Moles of water} + \text{Moles of nitrogen}} = \frac{7.14 \times 10^{-4}}{55.56 + 7.14 \times 10^{-4}} \approx 1.286 \times 10^{-5} \]

### Step 6: Apply Henry's Law
According to Henry's law, the relationship between the pressure of the gas and its mole fraction is given by:
\[ P_{N_2} = K_H \cdot X_{N_2} \]
Where:
- \(P_{N_2}\) is the pressure of nitrogen (1 bar),
- \(K_H\) is Henry's law constant,
- \(X_{N_2}\) is the mole fraction of nitrogen.

Rearranging the formula to find \(K_H\):
\[ K_H = \frac{P_{N_2}}{X_{N_2}} = \frac{1 \, \text{bar}}{1.286 \times 10^{-5}} \approx 77.7 \, \text{kbar} \]

### Final Result
The Henry's law constant \(K_H\) is approximately \(77.7 \, \text{kbar}\).

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