Q. The Henry’s law constant for the solubility of $N_{2}$ gas in water at 298 K is $1.0\ \times \ 105\ atm$. The mole fraction of $N_{2}$ In air is 0.8 The number of moles of $N_{2}$ from air dissolved in 10 moles of water of 298 K and 5 atm pressure is
A) $4\ \times \ 10^{- 4}$
B) $4.0\ \times 10^{- 5}$
C) $5.0\ \times 10^{- 4}$
D) $4.0\ \times 10^{- 6}$
Explanation
(a)
$P_{N_{2}} = K_{H}\ \times \ mole - fraction\ (N_{2})$
mole-fraction
$\left( N_{2} \right)\frac{1}{10^{5}} \times \ 0.8 \times 5\ = 4 \times 10^{- 5}mol^{- 1\ }$
In 10 mole solubility is $4 \times 10^{- 4}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.