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The Henry's law constant for the solubility of $N_2$ gas in water is $1.0\times 10^5$ atm.The mole fraction of $N_2$ in air is 0.8.The amount of $N_2$ from air dissolved in 10mol of water at 298K and 5 atm pressure is

$\begin{array}{1 1}4.0\times 10^{-4}mol\\4.0\times 10^{-5}mol\\5.0\times 10^{-4}mol\\4.0\times 10^{-6}mol\end{array} $

1 Answer

Answer : $4.0\times 10^{-4}mol$
Partial pressure of $N_2$ in 5 atm pressure,$p_{N_2}$=(0.8)(5 atm)=4 atm
By definition,Henry's law is $p_{N_2}=K_Hx_{N_2}$.
This gives $x_{N_2}=\large\frac{p_{N_2}}{K_H}=\frac{4\;atm}{1.0\times 10^5atm}=$$4.0\times 10^{-5}$
Since $x_{N_2}=n_{N_2}/(n_{N_2}+n_{H_2O})$,in 10 mol of water,the amount of $N_2$ will be given by the expression
$4.0\times 10^{-5}=\large\frac{n_{N_2}}{n_{N_2}+(10mol)}$
Solving for $n_{N_2}=\large\frac{4.0\times 10^{-5}(10mol)}{1-4.0\times 10^{-5}}$$=4.0\times 10^{-4}$ mol
answered Jul 23, 2014 by sreemathi.v
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