Question

The Henry's law constant for the solubility of $N_2$ gas in water at $298\text{K}$ is $1.0\times {10}^5\text{atm}$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ dissolved in $10$ moles of water at $298\text{K}$ and $5\text{atm}$ is $x\times {10}^{-4}$. Find the value of $x$.

Answer 1

By using formula,
$X_{N_2}=\frac{P_{N_2}}{K_H}$
Here, $K_H=\text{Henry's constant}$
$P_{N_2}=P_{air}\times \text{mole fraction of}{N}_2$
$X_{N_2}=\frac{5\times 0.8}{1\times {10}^5}$
$\Rightarrow X_{N_2}=4\times {10}^{-4}$
$\therefore \frac{n_{N_2}}{n_{N_2}+n_{H_{2}O}}=4\times {10}^{-5}$$=\frac{n_{N_2}}{n_{N_2}+10}$
$\because n_{N_2}<<<10$
$\therefore n=4\times {10}^{-4}$