The Henry's law constant for the solubility of `N_(2)` gas in water at `298 K` is `1.0 xx 10^(5) atm`. The mole fraction of `N_(2)` in air is `0.8`. The number of moles of `N_(2)` from air dissolved in `10` moles of water at `298 K` and `5 atm`. Pressure is:

Options

A. `4.0xx10^(-4)`
B. `4.0xx10^(-5)`
C. `5.0xx10^(-4)`
D. `4.0xx10^(-6)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Partial Pressure of N₂
We know that the mole fraction of N₂ in air is 0.8 and the total atmospheric pressure is 5 atm. According to Dalton's law of partial pressures, the partial pressure of a gas can be calculated as:

\[ P_{N_2} = \text{Mole fraction of } N_2 \times \text{Total pressure} \]

Substituting the values:

\[ P_{N_2} = 0.8 \times 5 \, \text{atm} = 4 \, \text{atm} \]

### Step 2: Use Henry's Law to Find the Mole Fraction of Dissolved N₂
Henry's law states that the partial pressure of a gas is equal to the Henry's law constant multiplied by the mole fraction of the gas in the solution:

\[ P_{N_2} = k_H \times X_{N_2} \]

Where:
- \( P_{N_2} \) = partial pressure of N₂ = 4 atm
- \( k_H \) = Henry's law constant = \( 1.0 \times 10^5 \, \text{atm} \)
- \( X_{N_2} \) = mole fraction of N₂ in the solution

Rearranging the equation to find \( X_{N_2} \):

\[ X_{N_2} = \frac{P_{N_2}}{k_H} = \frac{4 \, \text{atm}}{1.0 \times 10^5 \, \text{atm}} = 4 \times 10^{-5} \]

### Step 3: Relate Mole Fraction to Moles
The mole fraction \( X_{N_2} \) is defined as:

\[ X_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \]

Where:
- \( n_{N_2} \) = moles of N₂ dissolved
- \( n_{H_2O} \) = moles of water = 10 moles

Since \( n_{N_2} \) is very small compared to \( n_{H_2O} \), we can approximate:

\[ X_{N_2} \approx \frac{n_{N_2}}{10} \]

Substituting the value of \( X_{N_2} \):

\[ 4 \times 10^{-5} = \frac{n_{N_2}}{10} \]

### Step 4: Solve for \( n_{N_2} \)
Now, we can solve for \( n_{N_2} \):

\[ n_{N_2} = 10 \times 4 \times 10^{-5} = 4 \times 10^{-4} \, \text{moles} \]

### Final Answer
The number of moles of \( N_2 \) from air dissolved in 10 moles of water at 298 K and 5 atm is:

\[ \boxed{4 \times 10^{-4} \, \text{moles}} \]

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