At a certain temperature, the solubility of N_2 gas in water at 2.38 atm is 56.0 mg of N_2 gas/100 g of water. Calculate the solubility of N_2 gas in water, at the same temperature, if the partial pressure of N_2 gas over the solution is increased from 2.38 atm to 5.00 atm.

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To calculate the solubility of N2 gas in water, we can use Henry's Law, which states that the solubility of a gas is directly proportional to its partial pressure.

First, we need to determine the solubility constant, which is the proportionality constant in Henry's Law. We can find this value by rearranging the equation and solving for it.

Solubility = k * Partial Pressure

56.0 mg/100 g of water = k * 2.38 atm

Now, let's solve for k:

k = (56.0 mg/100 g) / 2.38 atm
k = 0.23529 mg/(g * atm)

Now that we have the solubility constant (k), we can use it to calculate the solubility when the partial pressure is increased to 5.00 atm.

Solubility = k * Partial Pressure

Solubility = 0.23529 mg/(g * atm) * 5.00 atm

Solubility = 1.17645 mg/g

Therefore, the solubility of N2 gas in water, at the same temperature, when the partial pressure is increased to 5.00 atm is approximately 1.17645 mg/g.

To calculate the solubility of N2 gas in water, you can use Henry's Law. According to Henry's Law, the solubility of a gas is directly proportional to its partial pressure.

Here's how you can calculate the solubility of N2 gas in water with the given information:

1. Convert the solubility provided from mg/100g to moles/100g. To do this, you need to know the molar mass of N2. The molar mass of N2 is 28.014 g/mol.

So, the solubility in moles/100g can be calculated as follows:
(56.0 mg N2/100 g water) / (28.014 g/mol) = 0.0019996 mol/100g water

2. Apply Henry's Law using the initial and final partial pressures of N2 gas.
According to Henry's Law, the solubility of a gas is directly proportional to its partial pressure. The general equation for Henry's Law is:

S₁/P₁ = S₂/P₂

Where:
S₁ and S₂ are the solubilities of the gas at the initial and final pressures, respectively.
P₁ and P₂ are the initial and final partial pressures of the gas, respectively.

In this case, we have:
S₁ = 0.0019996 mol/100g water (initial solubility)
P₁ = 2.38 atm (initial partial pressure)
P₂ = 5.00 atm (final partial pressure)

3. Rearrange the equation: S₂ = (S₁ × P₂) / P₁
Plug in the values:
S₂ = (0.0019996 mol/100g water × 5.00 atm) / 2.38 atm

4. Calculate the solubility (S₂):
S₂ ≈ 0.00839496 mol/100g water

5. Convert the obtained solubility from moles/100g to mg/100g. To do this, you can use the molar mass of N2. The molar mass of N2 is 28.014 g/mol.

Solubility in mg/100g = (0.00839496 mol/100g water) × (28.014 g/mol) = 0.23484 g/100g water
Converting g to mg: 0.23484 g/100g water × 1000 mg/g = 234.84 mg/100g water

Therefore, the solubility of N2 gas in water, when the partial pressure of N2 gas is increased from 2.38 atm to 5.00 atm, is approximately 234.84 mg/100g of water.

56 mg/100 mL = 560 mg/L = 0.560 g/L

Convert to mol/L = M = 0.560/molar mass N2.
partial pressure = kC
k = p/c = 2.38 atm/M of N2
moles N2 = 56 g/molar mass N2.
Find k, then use
p = kC, substitute 5.00 for p and k from above, solve for M N2 at 5.00 atm. The author of the problem may want you to show moles in the 100 mL so you may need to convert M to that.