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

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