Question

The Henry's law constant for the solubility $of{N}_2$ gas in water at $298K$ is $1.0\times {10}^5$ atm. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air dissolved in $10$ moles of water at $298K$ and $5$ atm pressure is:

• A. $4.0\times {10}^{-4}$
• B. $4.0\times {10}^{-5}$
• C. $5.0\times {10}^{-4}$
• D. $4.0\times {10}^{-6}$

Answer 1

The correct option is A $4.0\times {10}^{-4}$

According to Henry's law :

$P=K_H.X_{N_2}$

$0.8\times 5=1\times {10}^5\times X_{N_2}$

$X_{N_2}=4\times {10}^{-5}$ (in 10 moles ofwater)

$\Rightarrow 4\times {10}^{-5}=\frac{n_{N_2}}{n_{N_2}+10}$

$n_{N_2}\times 4\times {10}^{-5}+4\times {10}^{-4}=n_{N_2}$

$\Rightarrow n_{N_2}=4\times {10}^{-4}$

So, the correct option is $A$