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Questions  >>  CBSE XII  >>  Chemistry  >>  Solutions
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Q)

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} $

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A)
$P_{N_2}=$ 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 =$6.59\times 10^{7} mol L^{-1}$
but we are given henry,s constant in M/ mm of hg
so henry,s law should apply in the form :
conc. of solution=k(henry's constant for N2 ) * P(N2)(partial pressure of N2)
partial pressure of N2 =mol fraction of N2 in air * total pressure
P(n2)=0.78*1=0.78 atm
0.78*760=592.8 mm of hg
so, p(n2)=592.8
now ,
con. in sol.=Kh * P(n2)=(8.42 * 10^-7 )*592.8
                                    =4.99*10^-4 M
P(n2)= KH* x(n2)
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