Question

The Henry’s law constant for the solubility of $N_2$ gas in water at $298K$ is $1\times 10^{-5}atm.$ The mole fraction of $N_2$ in air is $0.8$ . If the number of moles of $N_2$ of air dissolved in $10moles$ of water at $298K$ and $5atm$ is $x\cdot 10^{-4}.$ Find the value of x.

Answer 1

Answer: 4

4.00

$X_{N_2}=\frac{P_{N_2}^1}{K_H}=\frac{5\times 0\cdot 8}{1\times 10^{-5}}$

$\left(P_{N_2}^1=P_{air}{\left(\text{mole fraction of N}\right)}_2\right)$

$=4\times 10^{-5}$

$\therefore \frac{{\text{n}}_{{\text{N}}_2}}{{\text{n}}_{{\text{N}}_2}+{\text{n}}_{{\text{H}}_2\text{O}}}=4\times 10^{-5}=\frac{{\text{n}}_{{\text{N}}_2}}{{\text{n}}_{{\text{N}}_2}+10}\sim eq\frac{{\text{n}}_{{\text{N}}_{\text{2}}}}{10}$

$\therefore \text{n}=4\times 10^{-4}$

Hence $x=4$