Question

Nitrogen, , 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 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

The right solution is "
`6.55* 10^(-3) \ g/100 \ ml`"

Step-by-step explanation:

The given values are:

Total pressure,

= 4.79 atm

Mole fraction,

= 78.1

Solubility,

S =
`1.75* 10^(-3)/100`

Partial pressure,

P = 1

By using the Henry's law,

⇒
`S=K* P`

On putting the given values, we get

⇒
`(1.75* 10^(-3))/(100) = K* 1`

⇒
`K=1.75* 10^(-5) g/ml \ atm` (Henry's constant)

The pressure of nitrogen (
`P'`) will be:

=
`Total \ pressure* Mole \ fraction`

On substituting the above given values, we get

=
`(4.79* 78.1 )/(100)`

=
`3.741 \ atm`

New solubility of nitrogen will be:

⇒
`S' = K* P'`

⇒
`=1.75* 10^(-5)* 3.741`

⇒
`=6.55* 10^(-5) \ g/ml`

So,

`The solubility of water will be:

=
`6.55* 10^(-5)* 100`

=
`6.55* 10^(-3) \ g/100 \ ml`