Answer:
Solubility of nitrogen in water at a nitrogen pressure of 4.5 atm is [tex]3.1\times 10^{-3}M[/tex]
Explanation:
According to Henry's law for solubility of a gas dissolved in a solvent-
[tex]\frac{C_{1}}{C_{2}}=\frac{P_{1}}{P_{2}}[/tex]
where [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are solubility of the gas at a pressure of [tex]P_{1}[/tex] and [tex]P_{2}[/tex] respectively.
Here, [tex]C_{1}=6.9\times 10^{-4}M[/tex], [tex]P_{1}=1.0atm[/tex] and [tex]P_{2}=4.5atm[/tex]
So, [tex]C_{2}=\frac{C_{1}P_{2}}{P_{1}}[/tex]
or, [tex]C_{2}=\frac{(6.9\times 10^{-4}M)\times (4.5atm)}{(1.0atm)}[/tex]
or, [tex]C_{2}=3.1\times 10^{-3}M[/tex]
So, solubility of nitrogen in water at a nitrogen pressure of 4.5 atm is [tex]3.1\times 10^{-3}M[/tex]
