Question

The mathematical relationship between gas solubility and pressure is called Henry's Law, solubility = kHPgas where kH is the Henry's Law constant for a gas at a given temperature. What is kH for Ar at 25 °C, in units of mol/L•mm Hg?

Answer 1

Answer : The unit of $k_H$ in mol/L.mm Hg is, $1.8\times 10^{-6}mol/L.mmHg$

Explanation :

As we know that the $k_H$ is the Henry's Law constant for argon at $25^oC$ is, $1.4\times 10^{-3}mol/L.atm$

Now we have to determine the unit of $k_H$ in mol/L.mm Hg

Conversion used for pressure from atm to mmHg is:

1 atm = 760 mmHg

So,

$k_H=1.4\times 10^{-3}mol/L.atm\times \frac{1atm}{760mmHg}$

$k_H=1.8\times 10^{-6}mol/L.mmHg$

Thus, the unit of $k_H$ in mol/L.mm Hg is, $1.8\times 10^{-6}mol/L.mmHg$