Question

The solubility of nitrogen gas in water at 25 °C and a nitrogen pressure of 1.0 atm is 6.9 × 10-4 M. The solubility of nitrogen in water at a nitrogen pressure of 4.5 atm is ________ M.

Answer 1

Solubility of nitrogen in water at a nitrogen pressure of 4.5 atm is
`3.1* 10^(-3)M`

Step-by-step explanation:

According to Henry's law for solubility of a gas dissolved in a solvent-

`(C_(1))/(C_(2))=(P_(1))/(P_(2))`

where
`C_(1)` and
`C_(2)` are solubility of the gas at a pressure of
`P_(1)` and
`P_(2)` respectively.

Here,
`C_(1)=6.9* 10^(-4)M`,
`P_(1)=1.0atm` and
`P_(2)=4.5atm`

So,
`C_(2)=(C_(1)P_(2))/(P_(1))`

or,
`C_(2)=((6.9* 10^(-4)M)* (4.5atm))/((1.0atm))`

or,
`C_(2)=3.1* 10^(-3)M`

So, solubility of nitrogen in water at a nitrogen pressure of 4.5 atm is
`3.1* 10^(-3)M`