Chapter 13, Problem 79GQ

The following table lists the concentrations of the principal ions in seawater:

1. (a) Calculate the freezing point of seawater.
2. (b) Calculate the osmotic pressure of seawater at 25 °C. What is the minimum pressure needed to purify seawater by reverse osmosis?

Interpretation Introduction

Interpretation: The freezing point of seawater has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Freezing point depression: The freezing point of the solution varies with the solute concentration.

$\begin{array}{c}\text{Freezing point depression = }\Delta {\text{T}}_{fp}{\text{= K}}_{fp}{\text{. m}}_{solute}\text{,}\\ \text{where,}{\text{K}}_{fp}\text{=}\text{molal freezing point depression constant,}\\ {\text{m}}_{solute}\text{= molality of solute}\text{.}\end{array}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 79GQ

Freezing point of seawater is $-{2.05}^{0}C$.

Explanation of Solution

Given,

  Molal freezing point depression constant of water is $-\text{1}{\text{.86}}^{\text{0}}\text{C/m}$

The value $1ppm$ means, it contains $1g$ of solute in ${10}^{6}g$ of solution.

Hence, the concentration given in ppm can be taken as the mass of each of the ions on ${10}^{6}g$ of water.

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

  $\begin{array}{c}NumberofmolesofC{l}^{-}=\frac{\text{1}\text{.95}\text{×}{\text{10}}^{\text{4}}\text{g}}{\text{35}\text{.45}\text{g/mol}}\\ =550.0mol\end{array}$

Number of moles of $C{l}^{-}$ is $550.0mol$

  $\begin{array}{c}NumberofmolesofN{a}^{+}=\frac{\text{1}\text{.08}\text{×}{\text{10}}^{\text{4}}\text{g}}{\text{22}\text{.99}\text{g/mol}}\\ =469.8mol\end{array}$

Number of moles of $N{a}^{+}$ is $469.8mol$

  $\begin{array}{c}NumberofmolesofM{g}^{+}=\frac{\text{1}\text{.29}\text{×}{\text{10}}^3\text{g}}{\text{24}\text{.31}\text{g/mol}}\\ =53.06mol\end{array}$

Number of moles of $M{g}^{+}$ is $53.06mol$

  $\begin{array}{c}NumberofmolesofS{O}_4^{3-}=\frac{\text{9}\text{.05}\text{×}{\text{10}}^2\text{g}}{\text{96}\text{.06}\text{g/mol}}\\ =9.421mol\end{array}$

Number of moles of $S{O}_4^{3-}$ is $9.421mol$

  $\begin{array}{c}NumberofmolesofC{a}^{2+}=\frac{\text{4}\text{.12×}{\text{10}}^2\text{g}}{\text{40}\text{.08}\text{g/mol}}\\ =10.28mol\end{array}$

Number of moles of $C{a}^{2+}$ is $10.28mol$

  $\begin{array}{c}Numberofmolesof{K}^{+}=\frac{\text{3}\text{.80}\text{×}{\text{10}}^2\text{g}}{39.10\text{g/mol}}\\ =9.719mol\end{array}$

Number of moles of $K^{+}$ is $9.719mol$

  $\begin{array}{c}NumberofmolesofB{r}^{-}=\frac{67\text{g}}{\text{79}\text{.90}\text{g/mol}}\\ =0.839mol\end{array}$

Number of moles of $B{r}^{-}$ is $0.839mol$

So the total moles of principle ions in ${10}^{6}g$ of seawater is $1103.1mol$

Molality of seawater is,

  $\begin{array}{c}\text{Molality (m) =}\frac{1103.1mol}{{\text{10}}^3\text{kg}}\\ =1.1031m\end{array}$

Depression in freezing point is,

  $\begin{array}{c}\Delta T_{fp}=K_{fp}{m}_{solute}\\ =\left(-\text{1}{\text{.86}}^{\text{0}}\text{C/m}\right)\times \left(1.1031m\right)\\ =-{2.05}^{0}C\end{array}$

Therefore,

Freezing point of seawater is,

  $\begin{array}{c}Fp=0^{0}C+\left(-{2.05}^{0}C\right)\\ =-{2.05}^{0}C\end{array}$

Freezing point of seawater is $-{2.05}^{0}C$.

Interpretation Introduction

Interpretation: The osmotic pressure of seawater at ${25}^{0}C$ has to be determined. Minimum pressure needed to purify the seawater using reverse osmosis method has to be calculated.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Osmotic pressure: The pressure created by the column of solution for the system at equilibrium is a measure of the osmotic pressure and is calculated by using the equation,

  $\pi =cRT$

  where,

  c is the molar concentration

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 79GQ

The osmotic pressure of seawater at ${25}^{0}C$ is $27atm$. Hence, there should be a minimum pressure of $27atm$ is needed to purify the seawater using reverse osmosis method

Explanation of Solution

Given,

  $\begin{array}{c}\text{R}\text{=}\text{0}\text{.082057}\text{L}{\text{.atm K}}^{-1}{\text{mol}}^{-1}\\ T={25}^{0}C=\left(25+273.2\right)K=298.2K\end{array}$

The molarity of the solute is $1.1031M$ since molality and molarity are same

The osmotic pressure of seawater is,

  $\begin{array}{c}\pi =cRT\\ =\left(1.1031M\right)\times \left(0.082057L.atm{K}^{-1}mo{l}^{-1}\right)\left(298.2K\right)\\ =27atm\end{array}$

The osmotic pressure of seawater at ${25}^{0}C$ is $27atm$

To purify the seawater using reverse osmosis method, there should be a minimum pressure of $27atm$ is needed.