Question

Henry's law constant for the solubility of $N_2$ gas in water at $298 \,K$ is $1.0 \times 10^5$ 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 $298\, K$ and $5$ atm pressure is

• A. $4.0 \times 10^{-4}$
• B. $4.0 \times 10^{-5}$
• C. $5.0 \times 10^{-4}$
• D. $4.0 \times 10^{-6}$

Answer 1

Answer: (A)

Total pressure is $5 atm$. The mole fraction of nitrogen is $0.8$.
Hence, the partial pressure of nitrogen $= P _{ T } \times X =0.8 \times 5=4 atm$
According to Henry's law, $P = KX$
$x$ is the mole fraction
K is the henry's law constant
$P$ is the pressure in atm.
Substitute values in the above expression.
$
X =\frac{4 atm }{1 \times 10^{5} atm }=4 \times 10^{-5}=\frac{ n }{10}
$
$
n =4.0 \times 10^{-4} mol
$