Question

\[K\] is Henry’s constant and has the unit:
A) $atm\,mo{l^{ - 1}}\,d{m^3}$.
B) $mo{l^{ - 1}}d{m^3}at{m^{ - 1}}$
C) $atm\,mol\,d{m^ - }^3$.
D) $mol\,d{m^ - }^3at{m^{ - 1}}$.

Answer 1

Hint:We know that, mathematically, Henry’s law is expressed as,
$X = {K_H}P$
Where $P$ is the pressure.
$X$ is solubility of the gas in liquid.
${K_H}$ is Henry’s law constant.

Complete step by step answer:
Let us see Henry’s law statement:
It states the amount of the given gas that dissolves in the liquid is directly proportional to the pressure of the gas in equilibrium with the liquid at constant temperature.
We know the unit of pressure P is \[atm\]and the unit of solubility of the gas at a fixed temperature\[mol/d{m^3}\].
To find the unit of the constant, Rewrite the Henry’s equation,
\[{K_H} = \dfrac{x}{P}\]
\[{K_H} = \dfrac{{mol/d{m^3}}}{{atm}}\]
Thus, the unit of Henry’s constant is $mol\,d{m^ - }^3at{m^{ - 1}}$.
Hence,
$\therefore $ Option D is correct.

Additional information:
Henry’s law applicable only in certain conditions. Now we discuss about that conditions as below,
It works only if the molecules are at equilibrium.
It does not work for gases at high pressure. For example, Nitrogen at high pressure it becomes soluble and it is harmful in the blood supply.
It does not work if there is a chemical reaction between the solute and solvent. For example, Hydrochloric acid reacts with water by a dissociation reaction to generate hydronium and chloride ions.

Note:
We must remember that the Henry’s constant and solubility of the gas in liquid (x) are inversely proportional which means Henry’s constant is greater for the gases with lower solubility. Solubility means the maximum quantity of solute dissolved in a certain quantity of a solution.