Q13.47P - Chemistry: Molecular Nature Of Matter And Change

Question

Argon makes up 0.93% by volume of air. Calculate its solubility (mol/L) in water at $\mathbf{20}\mathbf{°}\mathbf{C}$ and 1.0 atm. The Henry’s law constant for Ar under these conditions is  $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{mol}\mathbf{/}\mathbf{L}\mathbf{·}\mathbf{atm}$.

Step-by-Step Solution

Answer

The solubility of the mixture is proportional to the pressure of the gas. The solubility of the Argon gas will dissolve in the 0.93L of Air is $1.5\times {10}^{-3} {\mathrm{molL}}^{-1}$  at 1atm.

Definition

Solubility can be defined as the maximum number of moles of solute which can be dissolved in solvent up to its saturation point. Solubility defined the concentration of the solution. The equation of the solubility is given by Henry’s law which said that the solubility is directly proportional to the pressure of gas. As the pressure of the gas increase, the solubility increases and vice versa.

Mathematically, the Solubility of the gas is proportional to the pressure of the gas.

 

Henry’s Law,

$$\begin{gathered} \mathrm{Solubility}={\mathrm{K}}_{\mathrm{H}}\times {\mathrm{P}}_{\mathrm{gas}} \\ {\mathrm{K}}_{\mathrm{H}}=\mathrm{Henry}\text{'}\mathrm{s} \mathrm{Law} \mathrm{Constant} \\ {\mathrm{P}}_{\mathrm{gas}}=\mathrm{Pressure} \mathrm{of} \mathrm{gas} \end{gathered}$$

Numerical Explanation

Henry’s constant of the gas:

$$\begin{gathered} \mathrm{Henry}\text{'}\mathrm{s} \mathrm{constant} \mathrm{of} \mathrm{gas}=1.5\times {10}^{-3}\mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{atm}}^{-1} \\ \mathrm{Pressure} \mathrm{of} \mathrm{the} \mathrm{Ar} \mathrm{gas}=1 \mathrm{atm} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{Ar} \mathrm{gas}=0.93\% \\ \mathrm{Volume} \mathrm{of} \mathrm{Ar} \mathrm{gas} \mathrm{by} \mathrm{means} 0.93\mathrm{L} \mathrm{present} \mathrm{in} 100\mathrm{L} \mathrm{of} \mathrm{air} \mathrm{solution} \\ \mathrm{Molar} \mathrm{Mass} \mathrm{of} \mathrm{Ar} \mathrm{gas} = 40\mathrm{g}/\mathrm{mole} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{Henrys} \mathrm{law}, \mathrm{The} \mathrm{Pressure} \mathrm{of} \mathrm{the} \mathrm{gas} \mathrm{increases} \mathrm{the} \mathrm{so}lub\mathrm{ility} \mathrm{of} \mathrm{the} \mathrm{solute} \mathrm{increases}. \\ \mathrm{According} \mathrm{to} \mathrm{Henry}\text{'}\mathrm{s} \mathrm{Law}, \\ \mathrm{So}lub\mathrm{ility} = {\mathrm{K}}_{\mathrm{H}}\times {\mathrm{P}}_{\mathrm{gas}} \\ \mathrm{So}lub\mathrm{ility}=1.5\times {10}^{-3}\mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{atm}}^{-1}\times 1 \mathrm{atm} \\ \mathrm{So}lub\mathrm{ility}=1.5\times {10}^{-3}\mathrm{mol} {\mathrm{L}}^{-1} \end{gathered}$$

The pressure of the Ar gas 

$$\begin{gathered} \text{Pressure of the Ar gas}=1 \text{atm} \\ \text{Volume of the Ar gas}=0.93\% \\ \text{Volume of Ar gas bymeans 0.93L present in 100L of air solution} \\ \text{According to the Henrys law},\text{ The Pressure of the gas increases the solubility of the solute increases} \end{gathered}$$ 

The volume of the Ar gas  is calculated as follows:

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{Ar} \mathrm{gas}=0.93\% \\ \mathrm{Volume} \mathrm{of} \mathrm{Ar} \mathrm{gas} \mathrm{by} \mathrm{means} 0.93\mathrm{L} \mathrm{present} \mathrm{in} 100\mathrm{L} \mathrm{of} \mathrm{air} \mathrm{solution} \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{Ar} \mathrm{gas}=40\mathrm{g}/\mathrm{mole} \\ \mathrm{According} \mathrm{to} \mathrm{Henry}\text{'}\mathrm{s} \mathrm{Law}, \\ \mathrm{So}lub\mathrm{ility} = {\mathrm{K}}_{\mathrm{H}}\times {\mathrm{P}}_{\mathrm{gas}} \\ \mathrm{So}lub\mathrm{ility}=1.5\times {10}^{-3}\mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{atm}}^{-1}\times 1 \mathrm{atm} \\ \mathrm{So}lub\mathrm{ility}=1.5\times {10}^{-3}\mathrm{mol} {\mathrm{L}}^{-1}. \end{gathered}$$

The concentration of the gas is $1.5\times {10}^{-3}\text{mol} {\text{L}}^{-1}$ . The solubility of the mixture is proportional to the pressure of the gas. The solubility of the Argon gas dissolved in the 0.93L of Air at 1 atm.