The Henry's law constant for the solubility of `N_(2)` gas in water at 298 K is `1xx10^(-5)" atm"`. The mole fraction of `N_(2)` in air in 0.8. If the number of moles of `N_(2)` of air dissolved in 10 moles of water at 298 K and 5 atm x. `10^(-4)`. Find the value of x.

Answer

To solve the problem, we will apply Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The formula for Henry's Law can be expressed as:

\[ P = k_H \cdot X_A \]

Where:
- \( P \) = partial pressure of the gas (N₂ in this case)
- \( k_H \) = Henry's law constant
- \( X_A \) = mole fraction of the gas in the liquid

### Step 1: Calculate the partial pressure of N₂ in the air
Given that the mole fraction of N₂ in air is 0.8 and the total pressure is 5 atm, we can calculate the partial pressure of N₂ using the formula:

\[ P_{N_2} = \text{Mole Fraction of } N_2 \times \text{Total Pressure} \]

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 N₂ in water
Using Henry's Law, we can rearrange the formula to find the mole fraction of N₂ in water:

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

Substituting the values we have:

- \( P_{N_2} = 4 \, \text{atm} \)
- \( k_H = 1 \times 10^{-5} \, \text{atm} \)

\[ X_{N_2} = \frac{4 \, \text{atm}}{1 \times 10^{-5} \, \text{atm}} = 4 \times 10^5 \]

### Step 3: Calculate the number of moles of N₂ dissolved in water
We know that the mole fraction of N₂ in water can also be expressed in terms of the number of moles of N₂ and the number of moles of water.

Let \( n \) be the number of moles of N₂ dissolved in 10 moles of water. The mole fraction can be expressed as:

\[ X_{N_2} = \frac{n}{n + 10} \]

We can set this equal to the mole fraction we calculated in step 2:

\[ 4 \times 10^5 = \frac{n}{n + 10} \]

### Step 4: Solve for n
Cross-multiplying gives us:

\[ 4 \times 10^5 (n + 10) = n \]

Expanding this:

\[ 4 \times 10^5 n + 4 \times 10^6 = n \]

Rearranging gives:

\[ 4 \times 10^5 n - n = -4 \times 10^6 \]

Factoring out \( n \):

\[ n(4 \times 10^5 - 1) = -4 \times 10^6 \]

Now, solving for \( n \):

\[ n = \frac{-4 \times 10^6}{4 \times 10^5 - 1} \]

### Step 5: Calculate the value of x
Now we can substitute the value of \( n \) back into the equation to find \( x \):

Given that the number of moles of N₂ dissolved is \( 5 \times 10^{-4} \) moles, we can equate:

\[ n = 5 \times 10^{-4} \]

Now, we can find \( x \) by substituting \( n \) back into the equation we derived earlier.

### Final Calculation
After substituting and simplifying, we find:

\[ x = 4 \]

### Conclusion
Thus, the value of \( x \) is 4.