Question

The Henry’s law constant for the solubility of $N_2$ gas in water at $298\, K$ 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 $298\, K$ 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

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 $= 5\, atm$
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 $H_{2}O$, $n_{H_2O} = 10$
no. of nmoles of $N_{2}$, $n_{N_2 }= ?$
$\frac{n_{N_2}}{n_{N_2}+ n_{H_2O}} = x_{N_2}=4 \times 10^{-5}$
$\Rightarrow \frac{n_{N_2}}{10 +n_{N_2}} = 4\times10^{-5}$
$\Rightarrow n_{N_2} = 4 \times 10^{-4} \quad\left[\because n_{N_2} <<< 10\right]$