Question

The Henry’s law constant for the solubility of $N_{2}$ gas in water at $298 \, K$ 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 $10 \, moles$ of water at $298 \, K$ and $5 \, atm$ is $x\cdot 10^{- 4}.$ Find the value of x.

Answer 1

$X_{N_{2}}=\frac{P_{N_{2}}^{1}}{K_{H}}=\frac{5 \times 0 \cdot 8}{1 \times 10^{- 5}}$
$\left( P _{ N _{2}}^{1}= P _{\text {air }}\right.$ mole fraction of $\left.N _{2}\right)$
$=4\times 10^{- 5}$
$\therefore \frac{\text{n}_{\text{N}_{2}}}{\text{n}_{\text{N}_{2}} + \text{n}_{\text{H}_{2} \text{O}}} = 4 \times 10^{- 5} = \frac{\text{n}_{\text{N}_{2}}}{\text{n}_{\text{N}_{2}} + 10} \sim eq \frac{\text{n}_{\text{N}_{\text{2}}}}{10}$
$\therefore \text{n}=4\times 10^{- 4}$
Hence $x=4$