Chapter 7, Problem 7.72E

Estimate the osmotic pressure, freezing point, and boiling point of seawater, which you can approximate as equivalent to a $1.08\text{-molal}$ solution of $\text{NaCl}$. Use equations $7.47$ and $7.49$ to calculate $K_{\text{f}}$ and $K_{\text{b}}$ for ${\text{H}}_{\text{2}}\text{O}$, and use ${\Delta }_{\text{fus}}H$ $\left[{\text{H}}_{\text{2}}\text{O}\right]=6.009\text{ kJ/mol}$ and ${\Delta }_{\text{vap}}H\left[{\text{H}}_{\text{2}}\text{O}\right]=40.66\text{ kJ/mol}$. From what you know about seawater, what assumptions are we making?

Interpretation Introduction

Interpretation:

The osmotic pressure, freezing point, and boiling point of seawater are to be estimated. The value of $K_{\text{f}}$ and $K_{\text{b}}$ for ${\text{H}}_{\text{2}}\text{O}$ is to be calculated. The assumptions about seawater are to be stated.

Concept introduction:

Colligative properties are those properties of a material that depends on the number of particles of the substance present in material. The colligative properties are lowering of vapor pressure, elevation in boiling point, depression in freezing point and osmotic pressure.

Answer to Problem 7.72E

The osmotic pressure, freezing point, and boiling point of seawater are $52.5\text{ bar}$, $-4.01°\text{C}$ and $\text{101}\text{.1}°\text{C}$ respectively. The value of $K_{\text{f}}$ and $K_{\text{b}}$ for ${\text{H}}_{\text{2}}\text{O}$ is $1.858\text{K}\cdot \text{kg/mol}$ and $0.512\text{K}\cdot \text{kg/mol}$ respectively. It is assumed that only sodium chloride is present in the seawater, and no other salt or minerals are present in sea water.

Explanation of Solution

The molality of the seawater is $1.08\text{molal}$.

It is assumed that sea water only consists of water and sodium chloride.

The enthalpy of fusion of water is $6.009\text{ kJ/mol}$.

The enthalpy of vaporization of water is $40.66\text{ kJ/mol}$.

The molar mass of the water is $18\text{g/mol}$.

The boiling point of water is $373.15\text{K}$.

The freezing point of water is $273.15\text{K}$.

The sodium chloride salt gets dissociated in water to from sodium ion and chloride ion. Therefore, the number of particles into which sodium chloride is dissociated is $2$.

The formula of boiling point elevation constant is represented as,

$K_{\text{b}}=\frac{M_{\text{m}}R{T}_{\text{BP}}^2}{1000{\Delta }_{\text{vap}}H}$

Where,

• ${\Delta }_{\text{vap}}H$ represents the enthalpy of fusion of solvent.

• $M_{\text{m}}$ represents the molar mass of the solvent.

• $T_{\text{BP}}$ represents the boiling point of the solvent.

• $R$ represents the gas constant with value $8.314\text{ J/K}\cdot \text{mol}$.

Substitute the value of ${\Delta }_{\text{vap}}H$, $M_{\text{m}}$, $T_{\text{BP}}$ and $R$ in the above equation.

$\begin{array}{rll}{K}_{\text{b}}& =& \frac{\left(18\text{g/mol}\right)\left(8.314\text{ J/K}\cdot \text{mol}\right){\left(373.15\text{K}\right)}^2}{\left(1000\text{g/kg}\right)\left(40.66\text{ kJ/mol}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}\\ & =& 0.512\text{K}\cdot \text{kg/mol}\end{array}$

The formula for elevation in boiling point is given as,

$T_{\text{b, 2}}-T_{\text{b, 1}}=Nm{K}_{\text{b}}$

Where,

• $m$ represents the molality of the solute.

• $K_{\text{b}}$ represents the boiling point elevation constant.

• $T_{\text{b, 1}}$ represents the boiling point of the pure solvent.

• $T_{\text{b, 2}}$ represents the boiling point of the solution.

• $N$ represents the number of particles into which solute is dissociated.

Rearrange the equation for the value of $T_{\text{b, 2}}$.

$T_{\text{b, 2}}=Nm{K}_{\text{b}}+T_{\text{b, 1}}$

Substitute the value of, $N$$m$, boiling point of water and $K_{\text{b}}$ in the above equation.

$\begin{array}{rll}{T}_{\text{b, 2}}& =& \left(2\right)\left(1.08\text{molal}\right)\left(0.512\text{K}\cdot \text{kg/mol}\right)+373.15\text{K}\\ & =& \text{374}\text{.2559}\text{K}\end{array}$

The boiling point of the seawater in Kelvin is represented as,

$\begin{array}{rll}{T}_{\text{b, 2}}& =& \left(\text{374}\text{.2559}\text{K}-273.15\right)°\text{C}\\ & =& \text{101}\text{.1}°\text{C}\end{array}$

Therefore, the boiling point of the seawater is $\text{101}\text{.1}°\text{C}$.

The formula of freezing point depression constant is represented as,

$K_{\text{f}}=\frac{M_{\text{m}}R{T}_{\text{MP}}^2}{1000{\Delta }_{\text{fus}}H}$

Where,

• ${\Delta }_{\text{fus}}H$ represents the enthalpy of fusion of solvent.

• $M_{\text{m}}$ represents the molar mass of the solvent.

• $T_{\text{MP}}$ represents the melting point of the solvent.

• $R$ represents the gas constant with value $8.314\text{ J/K}\cdot \text{mol}$.

Substitute the value of ${\Delta }_{\text{fus}}H$, $M_{\text{m}}$, $T_{\text{MP}}$ and $R$ in the above equation.

$\begin{array}{rll}{K}_{\text{f}}& =& \frac{\left(18\text{g/mol}\right)\left(8.314\text{ J/K}\cdot \text{mol}\right){\left(273.15\text{K}\right)}^2}{\left(1000\text{g/kg}\right)\left(6.009\text{ kJ/mol}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}\\ & =& 1.858\text{K}\cdot \text{kg/mol}\end{array}$

The formula for depression in freezing point is given as,

$T_{\text{f, 1}}-T_{\text{f, 2}}=Nm{K}_{\text{f}}$ (1)

Where,

• $m$ represents the molality of the solute.

• $K_{\text{f}}$ represents the freezing point depression constant.

• $T_{\text{f, 1}}$ represents the freezing point of the pure solvent.

• $T_{\text{f, 2}}$ represents the freezing point of the solution.

Rearrange the equation for the value of $T_{\text{f, 2}}$.

$T_{\text{f, 2}}=T_{\text{f, 1}}-im{K}_{\text{f}}$

Substitute the value of $N$, $m$, freezing point of water and $K_{\text{f}}$ in the above equation.

$\begin{array}{rll}{T}_{\text{b, 2}}& =& 273.15\text{K}-\left(2\right)\left(1.08\text{molal}\right)\left(1.858\text{K}\cdot \text{kg/mol}\right)\\ & =& 269.1367\text{K}\end{array}$

The freezing point of the seawater in Kelvin is represented as,

$\begin{array}{rll}{T}_{\text{b, 2}}& =& \left(269.1367\text{K}-273.15\right)°\text{C}\\ & =& -4.01°\text{C}\end{array}$

Therefore, the freezing point of the seawater is $-4.01°\text{C}$.

The relation between mole fraction of a solute and molarity of the solution is given as,

$m_{\text{solute}}=\frac{1000\times x_{\text{solute}}}{\left(1-x_{\text{solvent}}\right)M_{\text{solvent}}}$

Where,

• $x_{\text{solute}}$ represents the mole fraction of solute.

• $M_{\text{solvent}}$ represents the molar mass of the solvent.

• $m_{\text{solute}}$ represents the molarity of the solute.

Rearrange the equation for the value of $x_{\text{solute}}$.

$x_{\text{solute}}=\frac{m_{\text{solute}}{M}_{\text{solvent}}}{1000+m_{\text{solute}}{M}_{\text{solvent}}}$

Substitute the value of molarity of seawater and molar mass of water in the above equation.

$\begin{array}{rll}{x}_{\text{solute}}& =& \frac{\left(1.08\text{molal}\right)\left(18\text{g/mol}\right)}{1000+\left(1.08\text{molal}\right)\left(18\text{g/mol}\right)}\\ & =& 0.0191\end{array}$

Molar volume of the water at standard temperature is $0.01801\text{L/mol}$. The molar volume of seawater is assumed equal to water.

The standard temperature is $298.15\text{K}$.

The osmotic pressure of the solution is given as,

$\Pi =\frac{N{x}_{\text{solute}}RT}{\overline{V}}$

Where,

• $\overline{V}$ represents the molar volume of the solution.

• $R$ represents the gas constant with value $0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}$.

• $T$ represents the temperature.

Substitute the value of $\overline{V}$, $R$, $T$, $x_{\text{solute}}$ and $N$ in the above equation.

$\begin{array}{rll}\Pi & =& \frac{\left(2\right)\left(0.0191\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(298.15\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 52.5\text{ bar}\end{array}$

The osmotic pressure of the seawater is $52.5\text{ bar}$.

Conclusion

The osmotic pressure, freezing point, and boiling point of seawater are $52.5\text{ bar}$, $-4.01°\text{C}$ and $\text{101}\text{.1}°\text{C}$ respectively. The value of $K_{\text{f}}$ and $K_{\text{b}}$ for ${\text{H}}_{\text{2}}\text{O}$ is $1.858\text{K}\cdot \text{kg/mol}$ and $0.512\text{K}\cdot \text{kg/mol}$ respectively. It is assumed that only sodium chloride is present in the seawater, and no other salt or minerals are present in sea water.