Question

Nitrogen, N2, is soluble in blood and can cause intoxication at sufficient concentration. For this reason, the U.S. Navy advises divers using compressed air not to go below 125 feet. The total pressure at this depth is 4.79 atm. If the solubility of nitrogen at 1.00 atm is 1.75E-3 g/100 mL of water, and the mole percent of nitrogen in air is 78.1, what is the solubility of nitrogen in water from air at 4.79 atm?

Answer 1

Answer – Given, depth = 125 foot, total pressure at this depth is 4.79 atm

solubility of nitrogen at 1.00 atm = 1.75*10^-3 g/100 mL , mole percent of nitrogen in air = 78.1

First we need to calculate the partial pressure of N₂

We know, mole percent of nitrogen in air = 78.1,so

100 % = 4.79 atm

So, 78.1 % = ?

= 3.74 atm

Now we need to use Henry's law

S = Kh*P

the solublity at g/mol

moles of N₂ = 1.75*10^-3 g / 28.014 g.mol^-1

                    = 6.24*10^-5 moles

     Solubility = 6.24*10^-5 moles / 0.100 L = 6.24*10^-4 M

Kh = S/P

      = 6.24*10^-4 M / 1.00 atm

       = 6.24*10^-4 M/atm

Now we have partial pressure of N₂ at 4.97 atm air

S = 6.24*10^-4 M/atm * 3.74 atm

    = 0.00234 M

Moles of N₂ = 0.00234 M *0.100 L = 0.000234 moles

So, solubility in g/mL

Mass of N₂ = 0.000234 moles * 28.014 g/mol

                  = 0.00655 g

So the solubility of nitrogen in water from air at 4.79 atm is 6.55*10^-3 g/100 mL of water.