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 $5atm$ 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

Answer: (A)

According to Henry’s law,
$x_{N_2}\times K_H=p_{N_2}$ ($p_{N_2}=$ Partial pressure of $N_2$)
Given, total pressure $=5atm$
mole fraction of $N_2=0.8$
$\therefore$ Partial pressure of $N_2=0.8\times 5=4$
$\Rightarrow x_{N_2}\times 1\times 10^5=4$
$\Rightarrow x_{N_2}=4\times 10^{-5}$
no. of moles of $H_{2}O$, $n_{H_{2}O}=10$
no. of nmoles of $N_2$, $n_{N_2}=?$
$\frac{n_{N_2}}{n_{N_2}+n_{H_{2}O}}=x_{N_2}=4\times 10^{-5}$
$\Rightarrow \frac{n_{N_2}}{10+n_{N_2}}=4\times 10^{-5}$
$\Rightarrow n_{N_2}=4\times 10^{-4}\left[\because n_{N_2}<<<10\right]$