Question

At a certain temperature, the solubility of N2 gas in water at 2.38atm is 56.0mg of N2 gas/100 g water . Calculate the solubility of N2 gas in water, at the same temperature, if the partial pressure of N2 gas over the solution is increased from 2.38atm to 5.00atm .

Answer 1

Using Henry's Law, the new solubility of N₂ gas in water at 5.00 atm is calculated to be 117.65 mg/100 g water, given that the solubility at 2.38 atm is 56.0 mg/100 g water.

Step-by-step explanation:

To calculate the solubility of N₂ gas in water when the partial pressure is increased, we can use Henry's Law, which states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The law is represented by the formula:

S₁ / P₁ = S₂ / P₂

where S₁ and S₂ are the solubilities of the gas at pressure P₁ and P₂, respectively.

Given that the solubility at 2.38 atm is 56.0 mg/100 g water, we can set up a proportion to find the new solubility (S₂) at 5.00 atm:

56.0 mg / 2.38 atm = S₂ / 5.00 atm

By solving for S₂:

S₂ = 56.0 mg * (5.00 atm / 2.38 atm)

S₂ = 117.65 mg (rounded to two decimal places)

Therefore, the new solubility at 5.00 atm is 117.65 mg of N₂ gas per 100 g of water.

Answer 2

Answer: The molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L

Step-by-step explanation:

We are given:

Solubility of nitrogen gas in water = 56.0 mg/100 g

Or, solubility of nitrogen gas in water = 0.056 g/100 mL (Density of water = 1 g/mL & Conversion factor used: 1 g = 1000 mg)

Solubility of a solute is defined as the moles of solute dissolved in 1 L of solvent.

Conversion factor used: 1 L = 1000 mL

Applying unitary method:

In 100 mL water, the amount of solute (nitrogen gas) dissolved is 0.056 grams

So, in 1000 mL of water, the amount of solute (nitrogen gas) dissolved will be =
`(0.056)/(100)* 1000=0.56g`

Converting this solubility into mol/L by dividing with the molar mass of nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

So, Solubility of nitrogen gas =
`(0.56g/L)/(28g/mol)=0.2mol/L`

To calculate the Henry's constant we use the equation given by Henry's law, which is:

`C_(N_2)=K_H* p_(N_2)` .........(1)

where,

`K_H` = Henry's constant

`C_(N_2)` = molar solubility of nitrogen gas = 0.02 mol/L

`p_(N_2)` = partial pressure of nitrogen gas = 2.38 atm

Putting values in equation 1, we get:

`0.02mol/L=K_H* 2.38atm\\\\K_H=(0.02mol/L)/(2.38atm)=8.40* 10^(-3)mol/L.atm`

When pressure is changed to 5.00 atm

Now,

`p_(N_2)=5.00atm\\\\K_H=8.40* 10^(-3)mol/L.atm`

Putting values in equation 1, we get:

`C_(N_2)=8.40* 10^(-3)mol/L.atm* 5.00atm=0.042mol/L`

Hence, the molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L