Chapter 7, Problem 7.86E

Determine osmotic pressures for each solution given. Assume complete dissociation of the solute.

(a) $10.0\text{g}$ of $\text{KBr}$ in $100.0\text{g}$ ${\text{H}}_2\text{O}$ at $\text{20}\text{.0}°\text{C}$

(b) $20.0\text{g}$ of ${\text{SrCl}}_2$ in $100.0\text{g}$ ${\text{H}}_2\text{O}$ at $\text{80}\text{.0}°\text{C}$

(c) $0.100\text{molal}$ $\text{HCl}$. at $\text{37}\text{.0}°\text{C}$ (This is a good approximation of stomach acid.)

Interpretation Introduction

(a)

Interpretation:

The osmotic pressure at $\text{20}\text{.0}°\text{C}$ 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 a colligative property and depends on the number of atoms of particle of the substance present in material.

Answer to Problem 7.86E

The osmotic pressure of the given solution is $40.58\text{ bar}$.

Explanation of Solution

The given temperature is $\text{20}\text{.0}°\text{C}$ and the composition of solution is $10.0\text{g}$ of $\text{KBr}$ and $100.0\text{g}$ of ${\text{H}}_2\text{O}$. The molar volume of water is $0.01801\text{L/mol}$.

The composition of the solution containing $10\text{g}$ of $\text{KBr}$ and $100.0\text{g}$ of ${\text{H}}_2\text{O}$ is calculated by the formula,

$x_1=\frac{n_1}{n_1+n_2}=\frac{\frac{{\text{W}}_{\text{1}}}{{\text{M}}_{\text{1}}}}{\frac{{\text{W}}_{\text{1}}}{{\text{M}}_{\text{1}}}\text{+}\frac{{\text{W}}_2}{{\text{M}}_2}}$

Where,

• $n_1$ and $n_2$ represents the number of moles of component 1 and 2 in the solution.

• ${\text{W}}_{\text{1}}$ and ${\text{W}}_2$ represents the given mass of component 1 and 2.

• ${\text{M}}_{\text{1}}$ and ${\text{M}}_2$ represents the molar mass of component 1 and 2.

Substitute the values of given mass and molar mass in the above formula to calculate the composition.

$\begin{array}{rll}{x}_1& =& \frac{\frac{10.0\text{g}}{119.0\text{g}}}{\frac{10.0\text{g}}{119.0\text{g}}\text{+}\frac{100.0\text{g}}{18.01\text{g}}}\\ & =& 0.015\end{array}$

The mole fraction of component 1 $x_1$ is $0.015$.

On complete dissociation of $\text{KBr}$, 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 values of $\overline{V}$, $R$, $T$, $x_{\text{solute}}$ or $x_1$ and $N$ in the above equation.

$\begin{array}{rll}\text{Π}& =& \frac{\left(2\right)\left(0.015\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(303\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 40.58\text{ bar}\end{array}$

The osmotic pressure of the given solution is $40.58\text{ bar}$.

Conclusion

The osmotic pressure of the given solution is $40.58\text{ bar}$.

Interpretation Introduction

(b)

Interpretation:

The osmotic pressure at $\text{80}\text{.0}°\text{C}$ 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 a colligative property and depends on the number of atoms of particle of the substance present in material.

Answer to Problem 7.86E

The osmotic pressure of the given solution is $107.55\text{ bar}$.

Explanation of Solution

The given temperature is $\text{80}\text{.0}°\text{C}$ and the composition of solution is $20.0\text{g}$ of ${\text{SrCl}}_2$ and $100.0\text{g}$ of ${\text{H}}_2\text{O}$. The molar volume of water is $0.01801\text{L/mol}$.

The composition of the solution containing $20.0\text{g}$ of ${\text{SrCl}}_2$ and $100.0\text{g}$ of ${\text{H}}_2\text{O}$ is calculated by the formula,

$x_1=\frac{n_1}{n_1+n_2}=\frac{\frac{{\text{W}}_{\text{1}}}{{\text{M}}_{\text{1}}}}{\frac{{\text{W}}_{\text{1}}}{{\text{M}}_{\text{1}}}\text{+}\frac{{\text{W}}_2}{{\text{M}}_2}}$

Where,

• $n_1$ and $n_2$ represents the number of moles of component 1 and 2 in the solution.

• ${\text{W}}_{\text{1}}$ and ${\text{W}}_2$ represents the given mass of component 1 and 2.

• ${\text{M}}_{\text{1}}$ and ${\text{M}}_2$ represents the molar mass of component 1 and 2.

Substitute the values of given mass and molar mass in the above formula to calculate the composition.

$\begin{array}{rll}{x}_1& =& \frac{\frac{20.0\text{g}}{158.53\text{g}}}{\frac{20.0\text{g}}{158.53\text{g}}\text{+}\frac{100.0\text{g}}{18.01\text{g}}}\\ & =& 0.022\end{array}$

The mole fraction of component 1 $x_1$ is $0.022$.

On complete dissociation of ${\text{SrCl}}_2$, 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.

$\begin{array}{rll}\text{Π}& =& \frac{\left(3\right)\left(0.022\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(303\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 107.55\text{ bar}\end{array}$

The osmotic pressure of the given solution is $107.55\text{ bar}$.

Conclusion

The osmotic pressure of the given solution is $107.55\text{ bar}$.

Interpretation Introduction

(c)

Interpretation:

The osmotic pressure at $\text{37}\text{.0}°\text{C}$ 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 a colligative property and depends on the number of atoms of particle of the substance present in material.

Answer to Problem 7.86E

The osmotic pressure of the given solution is $5.15\text{bar}$.

Explanation of Solution

The given temperature is $\text{37}\text{.0}°\text{C}$ and given solution is $0.100\text{molal}$ of $\text{HCl}$. The molar volume of water is $0.01801\text{L/mol}$.

The relationship between molality and mole fraction is,

$m_{\text{solute}}=\frac{1000\cdot x_{\text{solute}}}{x_{\text{solvent}}\cdot M_{\text{solvent}}}$

The term $x_{\text{solvent}}$ can be substituted with $\left(1-x_{\text{solute}}\right)$ as shown below.

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

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 molality of $\text{HCl}$ and molar mass of water in the above equation.

$\begin{array}{rll}{x}_{\text{solute}}& =& \frac{\left(0.100\text{molal}\right)\left(18\text{g/mol}\right)}{1000+\left(0.100\text{molal}\right)\left(18\text{g/mol}\right)}\\ & =& 0.0018\end{array}$

Thus, the value of $x_{\text{solute}}$ is $0.0018$.

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

The osmotic pressure of the solution is given as,

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

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 $i$ in the above equation.

$\begin{array}{rll}\text{Π}& =& \frac{\left(2\right)\left(0.0018\right)\left(0.08314\text{L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(310\text{K}\right)}{0.01801\text{L/mol}}\\ & =& 5.15\text{bar}\end{array}$

The osmotic pressure of the solution is $5.15\text{bar}$.

Conclusion

The osmotic pressure of the given solution is $5.15\text{bar}$.