Answer :
Answer: The molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L
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 = [tex]\frac{0.056}{100}\times 1000=0.56g[/tex]
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 = [tex]\frac{0.56g/L}{28g/mol}=0.2mol/L[/tex]
To calculate the Henry's constant we use the equation given by Henry's law, which is:
[tex]C_{N_2}=K_H\times p_{N_2}[/tex] .........(1)
where,
[tex]K_H[/tex] = Henry's constant
[tex]C_{N_2}[/tex] = molar solubility of nitrogen gas = 0.02 mol/L
[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas = 2.38 atm
Putting values in equation 1, we get:
[tex]0.02mol/L=K_H\times 2.38atm\\\\K_H=\frac{0.02mol/L}{2.38atm}=8.40\times 10^{-3}mol/L.atm[/tex]
When pressure is changed to 5.00 atm
Now,
[tex]p_{N_2}=5.00atm\\\\K_H=8.40\times 10^{-3}mol/L.atm[/tex]
Putting values in equation 1, we get:
[tex]C_{N_2}=8.40\times 10^{-3}mol/L.atm\times 5.00atm=0.042mol/L[/tex]
Hence, the molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L