constant for the solubility| Chemistry | JEE Advanced 2010

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The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at 298 K is $1.0 \times 10^5 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is 0.8 . The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

Select the correct option:

A

$4.0 \times 10^{-4}$

B

$4.0 \times 10^{-5}$

C

$5.0 \times 10^{-4}$

D

$4.0 \times 10^{-6}$

✓ Correct! Well done.

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