Chapter 6, Problem 9MCP

(a)

Interpretation Introduction

Interpretation:

The law describing dissolution of oxygen gas in water has to be explained.

Explanation of Solution

The law describing dissolution of oxygen gas in water is Henry’s law.

(b)

Interpretation Introduction

Interpretation:

Henry’s law has to be stated.

Explanation of Solution

Henry’s law states that at a particular temperature the number of moles of gas dissolved in a liquid is directly proportional to the pressure of that gas.  Also it can be stated as; solubility of the gas is directly proportional to the pressure exerted by that gas in the atmosphere that is contact with the liquid.

The mathematical expression for Henry’s law is given below.

    $M=kP$

Where $M$ is the concentration of gas dissolved in liquid, $P$ is the pressure of the gas and $k$ is a constant that depends on temperature.

(c)

Interpretation Introduction

Interpretation:

The equilibrium solubility of $O_2$ at $2atm$ and ${25}^{o}C$ has to be calculated.

Answer to Problem 9MCP

The equilibrium solubility of $O_2$ at $2atm$ and ${25}^{o}C$  is $2.6\times 10^{-3}mol/L.$

Explanation of Solution

Given that the Henry’s law constant $k$ for $O_2$ in aqueous solution is $1.3\times {10}^{-3}mol/\left(L.atm\right)$ and pressure is $2atm.$

The equilibrium solubility of $O_2$ can be calculated as follows,

    $\begin{array}{c}M=kP\\ \\ =\left(1.3\times {10}^{-3}mol/\left(L.atm\right)\right)\times 2atm\\ \\ =2.6\times 10^{-3}mol/L.\end{array}$

Solubility is $2.6\times 10^{-3}mol/L.$

(d)

Interpretation Introduction

Interpretation:

The solubility of oxygen gas in the solution in $\%\left(m/m\right)$ has to be calculated.

Concept introduction:

$\%\left(m/m\right)$ can be calculated as follows,

    $\%\left(m/m\right)=\frac{massofsolute}{massofsolution}\times 100\%$

Answer to Problem 9MCP

The solubility of oxygen gas in the solution in $\%\left(m/m\right)$ is $8.32\%.$

Explanation of Solution

Solubility obtained is $\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol/L}\text{.}$  That is, $\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol}$ is present in one liter solution.  From the number of moles amount of solute dissolved in gram can be calculated as follows,

Amount of solute $\text{=}\left(mol\right)\times \text{molar}\text{mass}\text{=}\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol}\times \text{32}\text{g/mol}\text{=}\text{0}\text{.0832g}$

Given that the density of solution is $1g/mL.$

So, total mass of one liter of solution will be $1g\left(volume\times density\right).$

Therefore, the solubility of oxygen gas in the solution in $\%\left(m/m\right)$ can be calculated as follows,

    $\begin{array}{c}\%\left(m/m\right)=\frac{massofsolute}{massofsolution}\times 100\%\\ \\ =\frac{0.0832g}{1g}\times 100\%\\ \\ =8.32\%.\end{array}$

Percent mass/mass is $8.32\%.$

(e)

Interpretation Introduction

Interpretation:

The solubility of oxygen gas in the solution in $ppt$ has to be calculated.

Concept introduction:

Concentration in $ppt$ can be calculated as follows,

    $ppt=\frac{massofsolute}{massofsolution}\times {10}^3$

Answer to Problem 9MCP

The solubility of oxygen gas in the solution in $ppt$ is $83.2ppt.$

Explanation of Solution

Solubility obtained is $\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol/L}\text{.}$  That is, $\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol}$ is present in one liter solution.  From the number of moles amount of solute dissolved in gram can be calculated as follows,

Amount of solute $\text{=}\left(mol\right)\times \text{molar}\text{mass}\text{=}\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol}\times \text{32}\text{g/mol}\text{=}\text{0}\text{.0832g}$

Given that the density of solution is $1g/mL.$

So, total mass of one liter of solution will be $1g\left(volume\times density\right).$

Therefore, the solubility of oxygen gas in the solution in $ppt$ can be calculated as follows,

    $\begin{array}{c}ppt=\frac{massofsolute}{massofsolution}\times {10}^3\\ \\ =\frac{0.0832g}{1g}\times {10}^3\\ \\ =83.2ppt.\end{array}$

Concentration in $ppt$ is $83.2ppt.$

(f)

Interpretation Introduction

Interpretation:

The solubility of oxygen gas in the solution in $ppm$ has to be calculated.

Concept introduction:

Concentration in $ppm$ can be calculated as follows,

    $ppm=\frac{massofsolute}{massofsolution}\times {10}^6$

Answer to Problem 9MCP

The solubility of oxygen gas in the solution in $ppm$ is $83200ppt.$

Explanation of Solution

Solubility obtained is $\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol/L}\text{.}$  That is, $\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol}$ is present in one liter solution.  From the number of moles amount of solute dissolved in gram can be calculated as follows,

Amount of solute $\text{=}\left(mol\right)\times \text{molar}\text{mass}\text{=}\text{2}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{mol}\times \text{32}\text{g/mol}\text{=}\text{0}\text{.0832g}$

Given that the density of solution is $1g/mL.$

So, total mass of one liter of solution will be $1g\left(volume\times density\right).$

Therefore, the solubility of oxygen gas in the solution in $ppm$ can be calculated as follows,

    $\begin{array}{c}ppm=\frac{massofsolute}{massofsolution}\times {10}^6\\ \\ =\frac{0.0832g}{1g}\times {10}^6\\ \\ =83200ppt.\end{array}$

Concentration in $ppm$ is $83200ppt.$

(g)

Interpretation Introduction

Interpretation:

Among $\%\left(m/m\right),pptandppm$, the one which is suitable for representing the concentration of the given solution has to be calculated.

Explanation of Solution

The given solution is not too dilute, so $\%\left(m/m\right)$ is a better way for expressing the concentration.  Since $\%\left(m/m\right)$ has a significant value, it can be used for concentration calculation.