Question

If $N_2$ gas is bubbled through water at $293K$, Calculate solubility of $N_2$ gas ? Assume that $N_2$ exerts a partial pressure of $0.987bar$. Given that Henry's law constant for $N_2$ at $293K$ is $76.48\frac{kbar}{mole/litre}$

• A. $1.29\times {10}^{-5}$$\frac{mole}{litre}$
• B. $1.29\times {10}^{-4}$$\frac{mole}{litre}$
• C. $2.9\times {10}^{-3}\frac{mole}{litre}$
• D. $2.9\times {10}^{-5}\frac{mole}{litre}$

Answer 1

Answer: (A)

$1.29\times {10}^{-5}$$\frac{mole}{litre}$

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.
$x=\frac{0.987bar}{76480bar}=1.29\times {10}^{-5}$ $\frac{mole}{litre}$