Answer: 7 milimoles of [tex]N_2[/tex] dissolve in 1 litre water.
Explanation:
Given: [tex]N_2[/tex] gas is passed through water at 293K.
Assumption :[tex]N_2[/tex] exerts a partial pressure of 0.987 bar.
Take : Henry's constant [tex]p_N_2[/tex] = 76.8 Kbar
[tex]K_H=76800[/tex]
To find : Number of moles of [tex]N_2[/tex] gas dissolve in 1 litre water.
According to Henry's law,
[tex]p_N_2=K_HX_{N_2}[/tex]
[tex]\Rightarrow\ 0.987=76800X_{N_2}[/tex]
[tex]\Rightarrow\ X_{N_2}=\dfrac{0.987}{76800}=0.0000128515625\approx1.29\times10^{-5}[/tex]
moles in 1 liter of water = [tex]\dfrac{1000}{18}=55.56\ [\text{Molar mass of }H_2O=18\ g][/tex]
Let n= moles of nitrogen
Total moles = 55.56+n
So,
[tex]X_{N_2}=\dfrac{n}{55.56+n}=1.29\times10^{-5}\\\\\Rightarrow\ 55.56\times1.29\times10^{-5}+1.29\times10^5n=n\\\\\Rightarrow\ n- 1.29\times10^5n=0.000714033\\\\\Rightarrow\ 0.999987n=0.000714033\\\\\Rightarrow\ n\approx7\text{ milimoles}[/tex]
Hence, 7 milimoles of [tex]N_2[/tex] dissolve in 1 litre water.
