Question

What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of ${25}^o$C, a total pressure of $1$ atmosphere and mole fraction of nitrogen in air of $0.78$. ($K_H$ for nitrogen $=8.42\times {10}^{-7}$M/mm Hg).

Answer 1

$p_{N_2}=1atm$

$p_{N_2}=K_H\times N_2$

By using Henry's Law

$x{N}_2=\frac{p_{N_2}}{K_H}=\frac{1}{8.42\times {10}^{-7}}$

$=\frac{n\left(N_2\right)}{n\left(H_{2}O\right)}$

$=x{N}_2.n\left(H_{2}O\right)$

$=x{N}_2.\frac{1000}{18}$

$=\frac{1}{8.42\times {10}^{-7}}\times 55.5$

$=6.59\times {10}^{7}moles$

Hence Nitrogen concentration

$=6.59\times {10}^{7}mol{L}^{-1}$