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Determine the solubility of $\ce{N2}$ in water exposed to air at $\pu{25^\circ C}$, if the atmospheric pressure is $\pu{1.2 atm}$, assume that the mole fraction of nitrogen is $0.78$ in air and the Henry's Law constant for nitrogen in water at this temperature is $\pu{6.1 \times 10^{-4} M/atm}$.

I have the answer, but i want to know how to solve for it. I've been reading through my notes and the book trying to find a reference through which to approach this problem and i have found none. If anyone can guide me in the right direction that would suffice.

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What are the units of your answer? You need a solubility, which is an amount per volume. If your amount is moles, the your solubility is in moles/liter: $\text{M}$.

That unit appears in the Henry's Law constant! $6.1\times 10^{-4} \frac{\text{M}}{\text{atm}}$. What can you do to get rid of the atmospheres in the denominator? You probably need the pressure and the mole fraction.

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Henry's law is given by the formula $p=K_\mathrm h\chi$. Here $p$ is partial pressure of the gas $K_\mathrm h$ is Henry's constant for that gas and $\chi$ is mole fraction.Using this you can find it.

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