Chapter 7, Problem 7.85E

Interpretation Introduction

(a)

Interpretation:

The osmotic pressure at $\text{30}\text{.0}°\text{C}$ for the given solution is to be calculated.

Concept introduction:

Osmotic pressure is defined as the minimum pressure applied on the solution to stop the flow of solvent molecules through the semi-permeable membrane.

Osmotic pressure is the colligative property and depends on the number of atoms of particle of the substance present in material.

Answer to Problem 7.85E

The osmotic pressure of the given solution is $279.75\text{ bar}$.

Explanation of Solution

Given temperature $\text{30}\text{.0}°\text{C}$ and the composition of solution is $0.100\text{mol}$ of $\text{NaCl}$ and $0.900\text{mol}$ of ${\text{H}}_2\text{O}$. The molar volume of water is $0.01801\text{L/mol}$.

The mole fraction of $\text{NaCl}$ calculated by the formula,

$x_{\text{solute}}=\frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

Where,

• $n_{\text{solute}}$ and $n_{\text{solvent}}$ represents the number of moles of solute and solvent in the solution.

Substitute the values of number of moles of each component in the above formula.

$\begin{array}{rll}{x}_1& =& \frac{0.100\text{mol}}{0.100\text{mol}+\text{0}\text{.900}\text{mol}}\\ & =& 0.1\end{array}$

The mole fraction of component $\text{NaCl}$$x_1$ is $0.1$.

On complete dissociation of $\text{NaCl}$, two species are obtained. Thus, the van’t Hoff factor $N$ is two.

The osmotic pressure of the solution is given as,

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

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.

$\begin{array}{rll}\text{Π}& =& \frac{\left(2\right)\left(0.1\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(303\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 279.75\text{ bar}\end{array}$

The osmotic pressure of the given solution is $279.75\text{ bar}$.

Conclusion

The osmotic pressure of the given solution is $279.75\text{ bar}$.

Interpretation Introduction

(b)

Interpretation:

The osmotic pressure at $\text{30}\text{.0}°\text{C}$ for the given solution is to be calculated.

Concept introduction:

Osmotic pressure is defined as the minimum pressure applied on the solution to stop the flow of solvent molecules through the semi-permeable membrane.

Osmotic pressure is the colligative property and depends on the number of atoms of particle of the substance present in material.

Answer to Problem 7.85E

The osmotic pressure of the given solution is $419.62\text{ bar}$.

Explanation of Solution

Given temperature $\text{30}\text{.0}°\text{C}$ and the composition of solution is $0.100\text{mol}$ of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $0.900\text{mol}$ of ${\text{H}}_2\text{O}$. The molar volume of water is $0.01801\text{L/mol}$.

The mole fraction of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ calculated by the formula,

$x_{\text{solute}}=\frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

Where,

• $n_{\text{solute}}$ and $n_{\text{solvent}}$ represents the number of moles of solute and solvent in the solution.

Substitute the values of number of moles of each component in the above formula.

$\begin{array}{rll}{x}_1& =& \frac{0.100\text{mol}}{0.100\text{mol}+\text{0}\text{.900}\text{mol}}\\ & =& 0.1\end{array}$

The mole fraction of component $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$$x_1$ is $0.1$.

On complete dissociation of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$, two species are obtained. Thus, the van’t Hoff factor $N$ is three.

The osmotic pressure of the solution is given as,

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

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.

$\begin{array}{rll}\text{Π}& =& \frac{\left(3\right)\left(0.1\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(303\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 419.62\text{ bar}\end{array}$

The osmotic pressure of the given solution is $419.62\text{ bar}$.

Conclusion

The osmotic pressure of the given solution is $419.62\text{ bar}$.

Interpretation Introduction

(c)

Interpretation:

The osmotic pressure at $\text{30}\text{.0}°\text{C}$ for the given solution is to be calculated.

Concept introduction:

Osmotic pressure is defined as the minimum pressure applied on the solution to stop the flow of solvent molecules through the semi-permeable membrane.

Osmotic pressure is the colligative property and depends on the number of atoms of particle of the substance present in material.

Answer to Problem 7.85E

The osmotic pressure of the given solution is $559.50\text{ bar}$.

Explanation of Solution

Given temperature $\text{30}\text{.0}°\text{C}$ and the composition of solution is $0.100\text{mol}$ of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ and $0.900\text{mol}$ of ${\text{H}}_2\text{O}$. The molar volume of water is $0.01801\text{L/mol}$.

The mole fraction of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ calculated by the formula,

$x_{\text{solute}}=\frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

Where,

• $n_{\text{solute}}$ and $n_{\text{solvent}}$ represents the number of moles of solute and solvent in the solution.

Substitute the values of number of moles of each component in the above formula.

$\begin{array}{rll}{x}_1& =& \frac{0.100\text{mol}}{0.100\text{mol}+\text{0}\text{.900}\text{mol}}\\ & =& 0.1\end{array}$

The mole fraction of component $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$$x_1$ is $0.1$.

On complete dissociation of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$, two species are obtained. Thus, the van’t Hoff factor $N$ is four.

The osmotic pressure of the solution is given as,

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

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.

$\begin{array}{rll}\text{Π}& =& \frac{\left(4\right)\left(0.1\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(303\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 559.50\text{ bar}\end{array}$

The osmotic pressure of the given solution is $559.50\text{ bar}$.

Conclusion

The osmotic pressure of the given solution is $559.50\text{ bar}$.