Q13.46 P - Chemistry: Molecular Nature Of Matter And Change

Question

The Henry’s law constant (k_H) for ${\mathbf{O}}_{\mathbf{2}}$  in water at ${\mathbf{20}}^{\mathbf{o}}\mathbf{C}$  in $\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{atm}}^{\mathbf{-}\mathbf{1}}$ .

(a) How many grams of  ${\mathbf{O}}_{\mathbf{2}}$ will dissolve in 2.50 L of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$  that is in contact with pure ${\mathbf{O}}_{\mathbf{2}}$  at 1.00 atm?

(b) How many grams of ${\mathbf{O}}_{\mathbf{2}}$ will dissolve in 2.50 L of  ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ that is in contact with air, where the partial pressure of is 0.209 atm?

Step-by-Step Solution

Answer

1. The mass of the   gas will dissolve in the 2.5L of  is 0.1024gat 1atm.
2. The mass of the   gas will dissolve in the 2.5L of  is 0.022gat 0.209atm.

Step 1: The grams of O 2 will dissolve in 2.50 L of that is in contact with pure O 2 at 1.00 atm.

Solubility is defined as how much a solvent can take in solute when dissolved.

 The Henry’s constant of gas  $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{atm}}^{\mathbf{-}\mathbf{1}}$

Pressure of the   gas is 1 atm.

Volume of the water is 2.5L.

Molar Mass of gas is 32g/mole.

 According to the Henry’s law, The Pressure of the gas increases the solubility of the solute increases.

 According to Henry’s Law.

$$\begin{gathered} \mathbf{Solubility}\mathbf{=}{\mathbf{K}}_{\mathbf{H}}\mathbf{\times }{\mathbf{P}}_{\mathbf{gas}} \\ \mathbf{Solubility}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{atm}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{1}\mathbf{atm} \\ \mathbf{Solubility}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

 The concentration of the gas is  $\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}$.

$$\begin{gathered} \mathbf{Moles} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{Concentration} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{\times }\mathbf{Volume} \mathbf{of} \mathbf{solution} \\ \mathbf{Moles} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{L} \\ \mathbf{Solubility}\mathbf{=}3.2\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol} \end{gathered}$$

 Number of moles of  ${\mathbf{O}}_{\mathbf{2}} \mathbf{gas}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol}$

Therefore, Mass of ${\mathbf{O}}_{\mathbf{2}}$:

$$\begin{gathered} \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{Moles} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{\times }\mathbf{Molar} \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2} \\ \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol}\mathbf{\times }\mathbf{32}\mathbf{mo} \\ \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1024}\mathbf{g} \end{gathered}$$

 The mass of the gas  $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1024}\mathbf{g}$.

Step 2: The grams of O 2 will dissolve in 2.50 L of H 2 O that is in contact with air, where the partial pressure of O 2 is 0.209 atm.

The Henry’s constant of gas  $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{atm}}^{\mathbf{-}\mathbf{1}}$

Pressure of the   gas is 0.209 atm.

Volume of the water is 2.5L.

Molar Mass of gas is 32g/mole.

 

As per Henry’s law, the pressure of the gas increases, the solubility of the solute increases.

 $$\begin{gathered} \mathbf{Solubility}\mathbf{=}{\mathbf{K}}_{\mathbf{H}}\mathbf{\times }{\mathbf{P}}_{\mathbf{gas}} \\ \mathbf{Solubility}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}{\mathbf{atm}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{209}\mathbf{atm} \\ \mathbf{Solubility}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}7\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

The concentration of the gas is $\mathbf{0}\mathbf{.}\mathbf{27}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}$ .

 $$\begin{gathered} \mathbf{Moles} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{Concentration} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{\times }\mathbf{Volume} \mathbf{of} \mathbf{solution}\backslash \mathrm{hfill} \\ \mathbf{Moles} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{27}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{molL}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{L} \\ \mathbf{Solubility}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{675}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol} \end{gathered}$$

Number of moles of  ${\mathbf{O}}_{\mathbf{2}} \mathbf{gas}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{675}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol}$

 

Therefore, the Mass of ${\mathbf{O}}_{\mathbf{2}}$:

 $$\begin{gathered} \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{Moles} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{\times }\mathbf{Molar} \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2} \\ \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{675}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{mol}\mathbf{\times }\mathbf{32}\mathbf{mol} \\ \mathbf{Mass} \mathbf{of} \mathbf{O}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{022}\mathbf{g} \end{gathered}$$

The mass of the gas  $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{022}\mathbf{g}\mathbf{.}$