Answer : The solubility of nitrogen in water at an atmospheric pressure will be, ![2.5125\times 10^{-4}mole/L](https://tex.z-dn.net/?f=2.5125%5Ctimes%2010%5E%7B-4%7Dmole%2FL)
Explanation :
First we have to calculate the partial pressure of nitrogen.
Formula used :
![p_{N_2}=X_{N_2}\times P_{atm}](https://tex.z-dn.net/?f=p_%7BN_2%7D%3DX_%7BN_2%7D%5Ctimes%20P_%7Batm%7D)
where,
= partial pressure of nitrogen = ?
= mole fraction of nitrogen = ![7.81\times 10^{-1}](https://tex.z-dn.net/?f=7.81%5Ctimes%2010%5E%7B-1%7D)
= atmospheric pressure = 0.480 atm
Now put all the given values in the above formula, we get :
![p_{N_2}=7.81\times 10^{-1}\times 0.480 atm](https://tex.z-dn.net/?f=p_%7BN_2%7D%3D7.81%5Ctimes%2010%5E%7B-1%7D%5Ctimes%200.480%20atm)
![p_{N_2}=0.375atm](https://tex.z-dn.net/?f=p_%7BN_2%7D%3D0.375atm)
Now we have to calculate the solubility of nitrogen in water.
Formula used :
![s_{N_2}=p_{N_2}\times K_H](https://tex.z-dn.net/?f=s_%7BN_2%7D%3Dp_%7BN_2%7D%5Ctimes%20K_H)
where,
= partial pressure of nitrogen = 0.375 atm
= solubility of nitrogen in water = ?
= Henry's constant = ![6.70\times 10^{-4}mole/L.atm](https://tex.z-dn.net/?f=6.70%5Ctimes%2010%5E%7B-4%7Dmole%2FL.atm)
Now put all the given values in the above formula, we get :
![s_{N_2}=0.375atm\times 6.70\times 10^{-4}mole/L.atm](https://tex.z-dn.net/?f=s_%7BN_2%7D%3D0.375atm%5Ctimes%206.70%5Ctimes%2010%5E%7B-4%7Dmole%2FL.atm)
![s_{N_2}=2.5125\times 10^{-4}mole/L](https://tex.z-dn.net/?f=s_%7BN_2%7D%3D2.5125%5Ctimes%2010%5E%7B-4%7Dmole%2FL)
Therefore, the solubility of nitrogen in water at an atmospheric pressure will be, ![2.5125\times 10^{-4}mole/L](https://tex.z-dn.net/?f=2.5125%5Ctimes%2010%5E%7B-4%7Dmole%2FL)