Animated Solution for Chemistry - Solutions: Henry's constant (in kbar) for four gases $\alpha ,\beta ,\gamma$ and $\delta$ in water at 298 K is given below : $\begin{array}{ccccc}& \alpha & \beta & \gamma & \delta \\ K_H& 50& 2& 2\times 10^{-5}& 0.5\end{array}$ (density of water $=10^3{\text{ kg m}}^{-3}$ at 298 K). This table implies that

Imagine you have a beaker of water and you are trying to dissolve different gases into it. Some gases dissolve easily, while others stubbornly stay out. What governs this behavior? Enter Henry's Law, a beautiful principle that connects the pressure of a gas to how much of it dissolves in a liquid.

Henry's Law states that the partial pressure of a gas ($p$) above a liquid is directly proportional to its mole fraction ($\chi$) in the liquid. Mathematically, it is expressed as:

Here, $K_H$ is the famous Henry's constant. It is unique for every gas and depends on the temperature.

Let's look closely at the equation. If we rearrange it to solve for the mole fraction (which represents solubility), we get:

This tells us something profound: at a constant pressure, the solubility of a gas is inversely proportional to its Henry's constant. A higher $K_H$ means the gas is less soluble.

Now, let's evaluate option (a). The table shows that gas $\alpha$ has the highest $K_H$ value ($50\text{ kbar}$). According to our inverse relationship, this means $\alpha$ must have the lowest solubility, not the highest. Therefore, option (a) is incorrect.

Option (b) claims that the solubility of gas $\gamma$ at $308\text{ K}$ is lower than at $298\text{ K}$.

Scientifically, this statement is generally true. As temperature increases, the kinetic energy of the dissolved gas molecules increases, allowing them to escape the liquid phase, thereby decreasing solubility.

However, there is a catch! The question specifically asks what this table implies. The provided table only contains data at a single temperature: $298\text{ K}$. We cannot logically deduce the temperature dependence of solubility from this isolated data set. Thus, option (b) is also incorrect in this context.

Now we turn our attention to options (c) and (d), which both mention a $55.5\text{ molal}$ solution. What exactly does this mean?

Molality is defined as the number of moles of solute per kilogram of solvent. So, a $55.5\text{ molal}$ aqueous solution contains exactly $55.5\text{ moles}$ of the gas dissolved in $1\text{ kg}$ ($1000\text{ g}$) of water.

To use Henry's Law, we need the mole fraction ($\chi$). First, let's find the number of moles of water in that $1\text{ kg}$:

This is a magical number in chemistry! Now, we can calculate the mole fraction of the gas:

Armed with the mole fraction ($\chi =0.5$), we can now test the remaining options. Let's check option (d) for gas $\delta$.

From the table, $K_H$ for $\delta$ is $0.5\text{ kbar}$. We need to be careful with units here. Since the options are in 'bar', let's convert $K_H$:

Now, plug this into Henry's Law:

This perfectly matches the statement in option (d)! The pressure of a $55.5\text{ molal}$ solution of $\delta$ is indeed $250\text{ bar}$. The mystery is solved, and the physics holds true.