Question

K is Henry’s constant and has unit:
(A) $atmmo{{l}^{-1}}d{{m}^{3}}$
(B) $mo{{l}^{-1}}d{{m}^{3}}at{{m}^{-1}}$
(C) $atmmold{{m}^{-3}}$
(D) $mold{{m}^{-3}}at{{m}^{-1}}$

Answer 1

Hint Henry’s law is the gas law composed of concentration and pressure in its formula. Thus, the unit will comprise of two specific units of concentration ($mold{{m}^{-3}}$) and pressure (atm) compulsorily.

Complete step by step solution:
Henry’s law is one of the gas laws formulated by Willian Henry in the early nineteenth century. The law is stated in various forms, two of them are;
(a) The partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of gas. The proportionality constant is the henry’s constant.
P $\propto$ x
P = $\left( K \right)x$
where, P is partial pressure of gas
x is mole fraction of gas
and K is Henry's constant.
(b) At constant temperature, the amount of given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of the gas in equilibrium with that liquid.
An equivalent way of stating the law is;
The solubility of a gas in liquid is directly proportional to the partial pressure of gas above the liquid.
C $\propto$ ${{P}_{gas}}$
C = $\left( K \right){{P}_{gas}}$
where,
C is the solubility of gas at fixed temperature in a particular solvent
K is the henry’s law constant
${{P}_{gas}}$is the partial pressure of the gas.
i.e.
(Solubility) = (Henry’s constant) (Partial pressure)
Now,
The unit of the Henry’s constant can be derived as;
The formulation-
C = (K) ${{P}_{gas}}$
C has the unit as ($mold{{m}^{-3}}$ )
and, ${{P}_{gas}}$ has the units as (atm)
Thus,
The unit of K can be calculated as,
K = $\dfrac{C}{{{P}_{gas}}}$
Therefore, the unit of K is $mold{{m}^{-3}}at{{m}^{-1}}$.

Hence, option (D) $mold{{m}^{-3}}at{{m}^{-1}}$ is correct.

Additional information:
There are various forms of Henry’s law discussed in the technical literature. Some of them are;
(i) Equation- K = $\dfrac{{{P}_{gas}}}{C}$
Here, K is called Henry's law volatility constant.
Dimensions- $d{{m}^{3}}atmmo{{l}^{-1}}$
(ii) Equation- K = $\dfrac{C}{{{P}_{gas}}}$
Here, K is called Henry's law solubility constant.
Dimension- $mold{{m}^{-3}}at{{m}^{-1}}$
(iii) Equation- K = $\dfrac{{{P}_{gas}}}{x}$
Dimension- $atmmo{{l}_{solution}}mo{{l}_{gas}}^{-1}$
(iv) Equation- K = $\dfrac{C}{{{C}_{gas}}}$
Dimension- dimensionless

Note: Many factors affect Henry's law constant but none of them can change the empirical formula for the law. So, do not get confused within the units.
The most common, efficient and prior most statement is about Henry's law solubility constant. So, the empirical unit of the Henry’s law constant is $mold{{m}^{-3}}at{{m}^{-1}}$ despite having many forms.