Answer : The unit of [tex]k_H[/tex] in mol/L.mm Hg is, [tex]1.8\times 10^{-6}mol/L.mmHg[/tex]
Explanation :
As we know that the [tex]k_H[/tex] is the Henry's Law constant for argon at [tex]25^oC[/tex] is, [tex]1.4\times 10^{-3}mol/L.atm[/tex]
Now we have to determine the unit of [tex]k_H[/tex] in mol/L.mm Hg
Conversion used for pressure from atm to mmHg is:
1 atm = 760 mmHg
So,
[tex]k_H=1.4\times 10^{-3}mol/L.atm\times \frac{1atm}{760mmHg}[/tex]
[tex]k_H=1.8\times 10^{-6}mol/L.mmHg[/tex]
Thus, the unit of [tex]k_H[/tex] in mol/L.mm Hg is, [tex]1.8\times 10^{-6}mol/L.mmHg[/tex]
