Question

$CO\left(g\right)$ is dissolved in $H_{2}O$ at ${25}^{o}C$ and 0.010 atm. Henry's law constant for this system is $5.80\times {10}^4$ atm. Thus, mole fraction of $CO\left(g\right)$ is

• A. $8.28\times {10}^{-7}$
• B. $0.01$
• C. $0.99$
• D. $1.72\times {10}^{-7}$

Answer 1

The correct option is D $1.72\times {10}^{-7}$

The mole fraction of a gas in solution is related to partial pressure of the gas above the solution by Henry's law.
$p_{CO}={\chi }_{CO}{K}_H$
$\therefore {\chi }_{CO}=\frac{p_{CO}}{K_H}=\frac{0.01atm}{5.8\times {10}^{4}atm}$
$=1.72\times {10}^{-7}$