$\overline{){\mathbf{C}}{\mathbf{=}}{\mathbf{kP}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{C}}{\mathbf{P}}{\mathbf{=}}\frac{\mathbf{k}\overline{)\mathbf{P}}}{\overline{)\mathbf{P}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathbf{k}}{\mathbf{=}}\frac{\mathbf{C}}{\mathbf{P}}{\mathbf{=}}\frac{\mathbf{8}\mathbf{.}\mathbf{21}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}{\displaystyle \frac{\mathbf{mol}}{\mathbf{L}}}}{\mathbf{0}\mathbf{.}\mathbf{790}\mathbf{}\mathbf{atm}}$

The solubility of nitrogen in water is 8.21 x 10 ^{-4} mol/L at 0°C when the N_{2} pressure above water is 0.790 atm. Calculate the Henry’s law constant for N_{2} in units of mol/L • atm for Henry’s law in the form *C* = *kP*, where *C* is the gas concentration in mol/L. Calculate the solubility of N_{2} in water when the partial pressure of nitrogen above water is 1.10 atm at 0°C.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.