Question

Henry’s law constant for the solubility of N₂ gas in water at 298K is 1.0 × 10⁵ atm. The mole fraction of N₂ in air is 0.8. The number of moles of N₂ from air dissolved in 10 moles of water at 298K and 5 atm pressure is

• A. 4.0 × 10^-4
• B. 4.0 × 10^-5
• C. 5.0 × 10^-4
• D. 4.0 × 10^-6

Answer 1

The Correct Option is A

We are given the following information:

• Henry's law constant for $N_2$ in water at 298 K, $K_H = 1.0 \times 10^5$ atm
• Temperature, T = 298 K
• Mole fraction of $N_2$ in air, $y_{N_2} = 0.8$
• Total pressure of air, $P_{total} = 5$ atm
• Number of moles of water, $n_{water} = 10$ moles

We need to find the number of moles of $N_2$ from air dissolved in 10 moles of water.

Step 1: Calculate the partial pressure of $N_2$ in the air.

According to Dalton's law of partial pressures, the partial pressure of a gas in a mixture is the product of its mole fraction and the total pressure.

$P_{N_2} = y_{N_2} \times P_{total}$

$P_{N_2} = 0.8 \times 5$ atm

$P_{N_2} = 4$ atm

Step 2: Apply Henry's law to find the mole fraction of $N_2$ dissolved in water.

Henry's law relates the partial pressure of a gas above a liquid to the mole fraction of the gas dissolved in the liquid:

$P_{N_2} = K_H \times x_{N_2}$

Where $x_{N_2}$ is the mole fraction of $N_2$ dissolved in water.

Rearranging the formula to solve for $x_{N_2}$:

$x_{N_2} = \frac{P_{N_2}}{K_H}$

$x_{N_2} = \frac{4 \text{ atm}}{1.0 \times 10^5 \text{ atm}}$

$x_{N_2} = 4 \times 10^{-5}$

Step 3: Calculate the number of moles of $N_2$ dissolved in water.

The mole fraction of $N_2$ in the solution is defined as:

$x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{water}}$

Where $n_{N_2}$ is the number of moles of dissolved $N_2$ and $n_{water}$ is the number of moles of water.

Since the solubility of $N_2$ in water is very low (indicated by the small mole fraction $x_{N_2} = 4 \times 10^{-5}$), the number of moles of dissolved $N_2$ ($n_{N_2}$) is much smaller than the number of moles of water ($n_{water}$). Therefore, we can approximate the denominator:

$n_{N_2} + n_{water} \approx n_{water}$

So, the mole fraction expression simplifies to:

$x_{N_2} \approx \frac{n_{N_2}}{n_{water}}$

Now, we can solve for $n_{N_2}$:

$n_{N_2} \approx x_{N_2} \times n_{water}$

$n_{N_2} \approx (4 \times 10^{-5}) \times (10 \text{ moles})$

$n_{N_2} \approx 40 \times 10^{-5}$ moles

$n_{N_2} \approx 4.0 \times 10^1 \times 10^{-5}$ moles

$n_{N_2} \approx 4.0 \times 10^{-4}$ moles

The final answer is \(4.0 \times 10^{-4}\).

Answer 2

1. Calculate the partial pressure of N₂ in the air:

According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is its mole fraction multiplied by the total pressure.

P_N2 = Mole fraction of N₂ × Total pressure

P_N2 = 0.8 × 5 atm = 4.0 atm

2. Apply Henry's Law to find the mole fraction of N₂ dissolved in water:

Henry's Law states: P_gas = K_H × X_{gas(dissolved)}

Where:

- P_gas is the partial pressure of the gas above the solution (P_N2 = 4.0 atm).
- K_H is Henry's law constant (1.0 × 10⁵ atm).
- X_{gas(dissolved)} is the mole fraction of the gas dissolved in the liquid (X_{N2(dissolved)}).
 

Rearranging the formula to solve for X_{N2(dissolved)}:

X_{N2(dissolved)} = P_N2 / K_H

X_{N2(dissolved)} = 4.0 atm / (1.0 × 10⁵ atm) = 4.0 × 10^-5

3. Calculate the number of moles of N₂ dissolved:

The mole fraction of N₂ dissolved is defined as:

X_{N2(dissolved)} = n_N2 / (n_N2 + n_water)

Where:

- n_N2 is the number of moles of N₂ dissolved.
- n_water is the number of moles of water (given as 10 moles).

So, 4.0 × 10^-5 = n_N2 / (n_N2 + 10)

Since the solubility of N₂ is very low (indicated by the small mole fraction), the number of moles of dissolved N₂ (n_N2) will be much smaller than the number of moles of water (n_water). Therefore, we can approximate the denominator: (n_N2 + n_water) ≈ n_water.

X_{N2(dissolved)} ≈ n_N2 / n_water

n_N2 ≈ X_{N2(dissolved)} × n_water

n_N2 ≈ (4.0 × 10^-5) × (10 moles)

n_N2 ≈ 40 × 10^-5 moles

n_N2 ≈ 4.0 × 10^-4 moles

Therefore, the number of moles of N₂ from air dissolved in 10 moles of water is 4.0 × 10^-4.

Final Answer: The final answer is $4.0 \times 10^{-4}$.