1 kg of water under a nitrogen pressure of 1 atmosphere dissolves 0.02 gm of nitrogen at 293 K. Calculate Henry' s law constant :

Options

A. 7.2 x `10^(-4)` atm
B. 7.7 x `10^(3)` atm
C. 2 x `10^(-5)` atm
D. 2 x `10^(-2)` atm

Answer

To calculate Henry's law constant (kH) for the given problem, we can follow these steps:

### Step 1: Understand the Given Information
- Mass of nitrogen (solute) = 0.02 g
- Mass of water (solvent) = 1 kg
- Pressure of nitrogen = 1 atmosphere
- Temperature = 293 K

### Step 2: Convert Mass of Nitrogen to Moles
To find the number of moles of nitrogen (N₂), we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of nitrogen (N₂) is approximately 28 g/mol.

\[ \text{Number of moles of N₂} = \frac{0.02 \text{ g}}{28 \text{ g/mol}} = 7.14 \times 10^{-4} \text{ moles} \]

### Step 3: Calculate Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, the mass of the solvent (water) is 1 kg.

\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{7.14 \times 10^{-4} \text{ moles}}{1 \text{ kg}} = 7.14 \times 10^{-4} \text{ mol/kg} \]

### Step 4: Apply Henry's Law
Henry's Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The formula is given by:
\[ C = k_H \cdot P \]
Where:
- \( C \) is the concentration (molality in this case),
- \( k_H \) is Henry's law constant,
- \( P \) is the pressure of the gas.

Rearranging the formula to find \( k_H \):
\[ k_H = \frac{C}{P} \]

### Step 5: Substitute Values into the Equation
Substituting the values we have:
- \( C = 7.14 \times 10^{-4} \text{ mol/kg} \)
- \( P = 1 \text{ atm} \)

\[ k_H = \frac{7.14 \times 10^{-4} \text{ mol/kg}}{1 \text{ atm}} = 7.14 \times 10^{-4} \text{ mol/(kg·atm)} \]

### Final Answer
The Henry's law constant \( k_H \) is approximately:
\[ k_H = 7.14 \times 10^{-4} \text{ mol/(kg·atm)} \]

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