Question

Henry's law constant for the solubility of $N_2$ gas in water at $298K$ 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 $298K$ 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 $5atm$. The mole fraction of nitrogen is $0.8$.
Hence, the partial pressure of nitrogen $=P_T\times X=0.8\times 5=4atm$
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.
$<br/>X=\frac{4atm}{1\times 10^{5}atm}=4\times 10^{-5}=\frac{n}{10}<br/>$
$<br/>n=4.0\times 10^{-4}mol<br/>$