Chapter 11, Problem 11.86QP

(a)

Interpretation Introduction

Interpretation: The osmotic pressure of each of the given aqueous solutions 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. The higher osmotic pressure containing solution has the higher concentration. Basically, with the help of the value of osmotic pressure the concentration of solvent is calculated.

To determine: The osmotic pressure of $10.0\text{g}$ of $\text{NaCl}$ in $1.50\text{L}$ of solution.

Answer to Problem 11.86QP

Solution

The osmotic pressure of $10.0\text{g}$ of $\text{NaCl}$ in $1.50\text{L}$ of solution is $\underset{\_}{5.566atm}$.

Explanation of Solution

Explanation

Given

The given mass of compound $\text{NaCl}$ is $10.0\text{g}$.

The given temperature is $27°\text{C}$.

The given volume of solution is $1.50\text{L}$.

The osmotic pressure $\left(\pi \right)$ of the given methanol 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 $27+273=300\text{K}$.

The molar mass of $\text{NaCl}$ is $58.44\text{g/mol}$.

The number of moles (n) of $\text{NaCl}$ is calculated by the formula,

$n=\frac{\text{Mass}\text{of}\text{NaCl}}{\text{Molar}\text{mass}}$

Substitute the values in the above formula.

$\begin{array}{l}n=\frac{\text{Mass}\text{of}\text{NaCl}}{\text{Molar}\text{mass}}\\ n=\frac{10.0\text{g}}{58.44\text{g/mol}}\\ n=0.17\text{mol}\end{array}$

Molarity of $\text{NaCl}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{NaCl}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values in the above formula.

$\begin{array}{l}\text{Molarity}=\frac{\text{Moles}\text{of}\text{NaCl}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}\\ \text{Molarity}=\frac{0.17\text{mol}}{1.50\text{L}}\\ \text{Molarity}=0.113\text{mol/L}\end{array}$

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.

For covalent compound the value of Van’t Hoff factor is equal one.

The Van’t Hoff factor for the given solution is two due to the ionic nature of $\text{NaCl}$.

Substitute the values in the above formula.

$\begin{array}{l}\pi =iMRT\\ \pi =2\times 0.113\text{mol/L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times \text{300}\text{K}\\ \pi =\underset{\_}{5.566atm}\end{array}$

(b)

Interpretation Introduction

To determine: The osmotic pressure of $10.0\text{mg/L}$ of ${\text{LiNO}}_3$.

Answer to Problem 11.86QP

Solution

The osmotic pressure of $10.0\text{mg/L}$ of ${\text{LiNO}}_3$ is $\underset{\_}{71.43atm}$.

Explanation of Solution

Explanation

Given

The given concentration of aqueous solution is $10.0\text{mg/L}$.

The given temperature is $300\text{K}$.

From the given value of concentration it is clear that the mass of compound ${\text{LiNO}}_3$ is $10\text{mg}$ that is $10\times {10}^{-3}\text{g}$.

The molar mass of the compound ${\text{LiNO}}_3$ is $68.94\text{g/mol}$.

The number of moles (n) of ${\text{LiNO}}_3$ is calculated by the formula,

$n=\frac{\text{Mass}\text{of}{\text{LiNO}}_3}{\text{Molar}\text{mass}}$

Substitute the values in the above formula.

$\begin{array}{l}n=\frac{\text{Mass}\text{of}{\text{LiNO}}_3}{\text{Molar}\text{mass}}\\ n=\frac{10\text{mg}}{68.94\text{g/mol}}\\ n=\frac{10\times {10}^{-3}\text{g}}{68.94\text{g/mol}}\\ n=1.45\text{mol}\end{array}$

Molarity of ${\text{LiNO}}_3$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Moles}\text{of}{\text{LiNO}}_3}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values in the above formula.

$\begin{array}{l}\text{Molarity}=\frac{\text{Moles}\text{of}{\text{LiNO}}_3}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}\\ \text{Molarity}=\frac{1.45\text{mol}}{1.00\text{L}}\\ \text{Molarity}=1.45\text{mol/L}\end{array}$

The Van’t Hoff factor for the given solution is three due to the ionic nature of the given compound.

Substitute the values in the above equation (1).

$\begin{array}{l}\pi =iMRT\\ \pi =2\times 1.45\text{mol/L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times \text{300}\text{K}\\ \pi =\underset{\_}{71.43atm}\end{array}$

(c)

Interpretation Introduction

To determine: The osmotic pressure of $0.222\text{M}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ (glucose).

Answer to Problem 11.86QP

Solution

The osmotic pressure of $0.222\text{M}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ (glucose) is $\underset{\_}{5.468atm}$.

Explanation of Solution

Explanation

Given

The given concentration of glucose is $0.222\text{M}$.

The given temperature is $300\text{K}$.

The Van’t Hoff factor for the given solution is one due to the nonionic nature of the given compound.

Substitute the values in the equation (1) given in part (a).

$\begin{array}{l}\pi =iMRT\\ \pi =1\times 0.222\text{mol/L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times \text{300}\text{K}\\ \pi =\underset{\_}{5.468atm}\end{array}$

(d)

Interpretation Introduction

To determine: The osmotic pressure of $0.00764\text{M}{\text{K}}_2{\text{SO}}_4$.

Answer to Problem 11.86QP

Solution

The osmotic pressure of $0.00764\text{M}{\text{K}}_2{\text{SO}}_4$ is $\underset{\_}{0.565atm}$.

Explanation of Solution

Explanation

Given

The given concentration of ${\text{K}}_2{\text{SO}}_4$ is $0.00764\text{M}$.

The given temperature is $300\text{K}$.

The Van’t Hoff factor for the given solution is three due to the ionic nature of the given compound.

Substitute the values in the equation (1) given in part (a).

$\begin{array}{l}\pi =iMRT\\ \pi =3\times 0.00764\text{mol/L}\times 0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\times \text{300}\text{K}\\ \pi =\underset{\_}{0.565atm}\end{array}$

Conclusion

1. a) The osmotic pressure of $10.0\text{g}$ of $\text{NaCl}$ in $1.50\text{L}$ of solution is $\underset{\_}{5.566atm}$.
2. b) The osmotic pressure of $10.0\text{mg/L}$ of ${\text{LiNO}}_3$ is $\underset{\_}{71.43atm}$.
3. c) The osmotic pressure of $0.222\text{M}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ (glucose) is $\underset{\_}{5.468atm}$.
4. d) The osmotic pressure of $0.00764\text{M}{\text{K}}_2{\text{SO}}_4$ is $\underset{\_}{0.565atm}$.