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Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Solutions
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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} $

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