Chapter 11, Problem 11.104AP

Interpretation Introduction

Interpretation: The given one hundred milliliters of $2.50\text{mM}\text{NaCl}$ is being mixed with $80.0\text{mL}$ of $3.60\text{mM}{\text{MgCl}}_2$ at $20°\text{C}$. The osmotic pressure of each starting solution and that of the mixture is to be calculated.

Concept introduction: Semipermeable membrane is a thin film which is present almost in both living and nonliving things. This film comprises small pores through which solvent moves from one portion to another. Solvent always flows from lower concentration to higher concentration. This process of solvent flow is known as osmosis. Now if another side external pressure is applied then this process reversed and this applied reverse pressure is known as osmotic pressure.

To determine: The osmotic pressure of each starting solution and the mixture.

Answer to Problem 11.104AP

Solution

The osmotic pressure of the $\text{NaCl}$ solution at $20°\text{C}$ is $\underset{\_}{0.120atm}$.

The osmotic pressure of the ${\text{MgCl}}_2$ solution at $20°\text{C}$ is $\underset{\_}{0.259atm}$.

The osmotic pressure of the mixture solution at $20°\text{C}$ is $\underset{\_}{0.1821atm}$.

Explanation of Solution

Explanation

Given

The given amount of $\text{NaCl}$ solute is $2.50\text{mM}$.

The given amount of ${\text{MgCl}}_2$ solute is $3.60\text{mM}$.

The given volume of solvent for $\text{NaCl}$ is $100\text{mL}$.

The given volume of solvent for ${\text{MgCl}}_2$ is $80.0\text{mL}$.

The given temperature is $20°\text{C}$.

The osmotic pressure $\left(\pi \right)$ for solution is calculated by the formula,

$\pi =iMRT$ (1)

Where,

• $i$ is the van’t Hoff factor.
• $M$ is the molarity of the solution.
• $R$ is the gas constant.
• $T$ is t he temperature.

The value of gas constant is $0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}$.

The value of temperature in Kelvin is $20+273=293\text{K}$.

Van’t Hoff factor is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

Colligative properties are basically the properties of the solutions after the process of dissolution. It is already defined above that the vapor pressure of solvent changes after the dissolution.

For ionic compound the value of Van’t Hoff factor is equal to the number of ions.

Therefore, for the $\text{NaCl}$ compound the value of Van’t Hoff factor is two.

For the ${\text{MgCl}}_2$ compound the value of Van’t Hoff factor is three.

The molarity (M) of the given solution is calculated by the formula,

$M=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solvent}\text{(L)}}$ (2)

Starting solutions are the individual solutions of both $\text{NaCl}$ and ${\text{MgCl}}_2$ compound.

Therefore, calculations for the first solution that is solution of $\text{NaCl}$,

The molarity of $\text{NaCl}$ is $2.50\text{mM}$ that is $2.50\times {10}^{-3}\text{M}$.

The volume of solution in liter is $100\text{mL}$.

Substitute the values in equation (1).

$\begin{array}{l}\pi =iMRT\\ \pi =2\times 2.50\times {10}^{-3}\text{mol}/\text{L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times 293\text{K}\\ \pi =\underset{\_}{0.120atm}\end{array}$

Calculations for the second solution that is solution of ${\text{MgCl}}_2$,

The molarity of ${\text{MgCl}}_2$ is $3.60\text{mM}$ that is $3.60\times {10}^{-3}\text{M}$.

The volume of solution in liter is $80\text{mL}$.

Substitute the values in equation (1).

$\begin{array}{l}\pi =iMRT\\ \pi =3\times 3.60\times {10}^{-3}\text{mol}/\text{L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times 293\text{K}\\ \pi =\underset{\_}{0.259atm}\end{array}$

Now, calculations for the mixture,

When both the individual solutions are mixed with each other, the concentrations of both the solutions changed.

The total volume of the mixture is $100\text{mL}+80\text{mL}=180\text{mL}$.

The new moles of $\text{NaCl}$ is calculated by the formula,

$\text{Moles}=\text{Volume}\left(\text{L}\right)\times \text{Molarity}$

Substitute the values in the above formula,

$\begin{array}{l}\text{Moles}=\text{Volume}\left(\text{L}\right)\times \text{Molarity}\\ \text{Moles}=0.100\text{L}\times 2.50\times {10}^{-3}\text{M}\\ \text{Moles}=2.50\times {10}^{-4}\text{mol}\end{array}$

New concentration in terms of molarity for $\text{NaCl}$ is calculated by the formula given in equation (1).

$\begin{array}{l}M=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solvent}\text{(L)}}\\ M=\frac{2.50\times {10}^{-4}\text{mol}}{0.180\text{L}}\\ M=13.88\times {10}^{-4}\text{mol}/\text{L}\\ M=13.88\times {10}^{-4}\text{M}\end{array}$

Calculation for ${\text{MgCl}}_2$,

The new moles of ${\text{MgCl}}_2$ is calculated by the formula,

$\text{Moles}=\text{Volume}\left(\text{L}\right)\times \text{Molarity}$

Substitute the values in the above formula,

$\begin{array}{l}\text{Moles}=\text{Volume}\left(\text{L}\right)\times \text{Molarity}\\ \text{Moles}=0.080\text{L}\times 3.60\times {10}^{-3}\text{M}\\ \text{Moles}=2.88\times {10}^{-4}\text{mol}\end{array}$

New concentration in terms of molarity for ${\text{MgCl}}_2$ is calculated by the formula given in equation (1).

$\begin{array}{l}M=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solvent}\text{(L)}}\\ M=\frac{2.88\times {10}^{-4}\text{mol}}{0.180\text{L}}\\ M=16.00\times {10}^{-4}\text{mol}/\text{L}\\ M=16.00\times {10}^{-4}\text{M}\end{array}$

Osmotic pressure of the mixture is the sum of the osmotic pressure of individual solutions. It is expressed as,

$\begin{array}{l}{\pi }_{\text{mixture}}={\left(iMRT\right)}_{\text{NaCl}}+{\left(iMRT\right)}_{{\text{MgCl}}_2}\\ {\pi }_{\text{mixture}}={\pi }_{\text{NaCl}}+{\pi }_{{\text{MgCl}}_2}\end{array}$ (3)

Calculation for $\text{NaCl}$

Substitute the values for $\text{NaCl}$,

$\begin{array}{l}{\pi }_{\text{NaCl}}={\left(iMRT\right)}_{\text{NaCl}}\\ {\pi }_{\text{NaCl}}=2\times 13.88\times {10}^{-4}\text{mol}/\text{L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times 293\text{K}\\ {\pi }_{\text{NaCl}}=\text{0}\text{.0667}\text{atm}\end{array}$

Calculation for ${\text{MgCl}}_2$

Substitute the values for ${\text{MgCl}}_2$,

$\begin{array}{l}{\pi }_{{\text{MgCl}}_2}={\left(iMRT\right)}_{{\text{MgCl}}_2}\\ {\pi }_{{\text{MgCl}}_2}=3\times 16.00\times {10}^{-4}\text{mol}/\text{L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times 293\text{K}\\ {\pi }_{{\text{MgCl}}_2}=\text{0}\text{.1154}\text{atm}\end{array}$

Now, substitute the calculated values in equation (3).

$\begin{array}{l}{\pi }_{\text{mixture}}={\pi }_{\text{NaCl}}+{\pi }_{{\text{MgCl}}_2}\\ {\pi }_{\text{mixture}}=\text{0}\text{.0667}\text{atm}+\text{0}\text{.1154}\text{atm}\\ {\pi }_{\text{mixture}}=\underset{\_}{0.1821atm}\end{array}$

Conclusion

The osmotic pressure of the $\text{NaCl}$ solution at $20°\text{C}$ is $\underset{\_}{0.120atm}$.

The osmotic pressure of the ${\text{MgCl}}_2$ solution at $20°\text{C}$ is $\underset{\_}{0.259atm}$.

The osmotic pressure of the mixture solution at $20°\text{C}$ is $\underset{\_}{0.1821atm}$.