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

$\begin{array}{1 1}6.59\times 10^7mol L^{-1}\\5.59\times 10^7mol L^{-1}\\4.59\times 10^7mol L^{-1}\\3.59\times 10^7mol L^{-1}\end{array}$

$P_{N_2}=$ 1 atmosphere
$P_{N_2}=K_H\times N_2$
By Henry's law
$xN_2=\large\frac{P_{N_2}}{K_H}=\frac{1}{8.42\times 10^{-7}}$
$x{N_2}=\large\frac{n(N_2)}{n(N_2)+n(H_2O)}$
$xN_2=\large\frac{n(N_2)}{n(H_2O)}$
$n(N_2)=xN_2.n(H_2O)$
$nN_2=xN_2.\large\frac{1000}{18}$
$\Rightarrow \large\frac{1}{8.42\times 10^{-7}}$$\times 55.5$
$\Rightarrow 6.59\times 10^7$ moles
Hence concentration of nitrogen =$6.59\times 10^{7} mol L^{-1}$