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

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 = $\frac{0.056}{100}\times 1000=0.56g$

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 = $\frac{0.56g/L}{28g/mol}=0.2mol/L$

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

$C_{N_2}=K_H\times p_{N_2}$       .........(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:

$0.02mol/L=K_H\times 2.38atm\\\\K_H=\frac{0.02mol/L}{2.38atm}=8.40\times 10^{-3}mol/L.atm$

When pressure is changed to 5.00 atm

Now,

$p_{N_2}=5.00atm\\\\K_H=8.40\times 10^{-3}mol/L.atm$

Putting values in equation 1, we get:

$C_{N_2}=8.40\times 10^{-3}mol/L.atm\times 5.00atm=0.042mol/L$

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