<u>Answer:</u> The molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L
<u>Explanation:</u>
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 = ![\frac{0.056}{100}\times 1000=0.56g](https://tex.z-dn.net/?f=%5Cfrac%7B0.056%7D%7B100%7D%5Ctimes%201000%3D0.56g)
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 = ![\frac{0.56g/L}{28g/mol}=0.2mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B0.56g%2FL%7D%7B28g%2Fmol%7D%3D0.2mol%2FL)
To calculate the Henry's constant we use the equation given by Henry's law, which is:
.........(1)
where,
= Henry's constant
= molar solubility of nitrogen gas = 0.02 mol/L
= partial pressure of nitrogen gas = 2.38 atm
Putting values in equation 1, we get:
![0.02mol/L=K_H\times 2.38atm\\\\K_H=\frac{0.02mol/L}{2.38atm}=8.40\times 10^{-3}mol/L.atm](https://tex.z-dn.net/?f=0.02mol%2FL%3DK_H%5Ctimes%202.38atm%5C%5C%5C%5CK_H%3D%5Cfrac%7B0.02mol%2FL%7D%7B2.38atm%7D%3D8.40%5Ctimes%2010%5E%7B-3%7Dmol%2FL.atm)
<u>When pressure is changed to 5.00 atm</u>
Now,
![p_{N_2}=5.00atm\\\\K_H=8.40\times 10^{-3}mol/L.atm](https://tex.z-dn.net/?f=p_%7BN_2%7D%3D5.00atm%5C%5C%5C%5CK_H%3D8.40%5Ctimes%2010%5E%7B-3%7Dmol%2FL.atm)
Putting values in equation 1, we get:
![C_{N_2}=8.40\times 10^{-3}mol/L.atm\times 5.00atm=0.042mol/L](https://tex.z-dn.net/?f=C_%7BN_2%7D%3D8.40%5Ctimes%2010%5E%7B-3%7Dmol%2FL.atm%5Ctimes%205.00atm%3D0.042mol%2FL)
Hence, the molar solubility of nitrogen gas when pressure is increased is 0.042 mol/L