Chapter 13, Problem 139AP

Lysozyme is an enzyme that cleaves bacterial cell walls. A sample of lysozyme extracted from egg white has a molar mass of 13,930 g. A quantity of 0.100 g of this enzyme is dissolved in 150 g of water at $\text{25°C}\text{.}$ Calculate the vapor-pressure lowering, the depression in freezing point, the elevation in boiling point, and the osmotic pressure of this solution. (The vapor pressure of water at $\text{25°C}$ is 23.76 mmHg.)

Interpretation Introduction

Interpretation:

The vapor pressure lowering, depression in freezing point, elevation in boiling point, and atmospheric pressure of the solution are to be calculated with given molar mass of enzyme.

Concept Introduction:

The property that depends on the concentration of solute molecules and not on the identity of solute is termed as colligative property.

The four colligative properties of solution are vapor-pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Vapor pressure lowering is directly proportional to the mole fraction of the solute.

The pressure exerted by a solvent is equal to the product of that solvent’s mole fraction and the pressure exerted by the pure solvent. It is expressed as

$P_1={\chi }_1{P}_1{}^{\circ }$

Here, $P_1{}^{\circ }$ is the vapor pressure of the pure solvent, $P_1$ is the pressure exerted by a solvent over a solution, and ${\chi }_1$ is the mole fraction of the solvent.

Freezing point depression occurs when a nonvolatile solute is added to a solvent.

Freezing point depression is calculated by using the expression given below:

$\Delta T=K_f.m$

Here, $K_f$ is the freezing point constant and $m$ is the molality.

Elevation in boiling point occurs when a nonvolatile solute is added to a pure solvent.

Boiling point elevation is calculated by using the expression given below:

$\Delta T=K_b.m$

Here, $K_b$ is the boiling point constant and $m$ is the molality.

Osmotic pressure is the pressure that has to be applied to prevent the pure solvent to pass to a solution by osmosis.

The osmotic pressure is calculated by using the expression given below:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, $M$ is the molarity, $\text{R}$ is the gas constant and $T$ is the temperature.

The number of moles is calculated by dividing the given mass of the compound by the molar mass. It is expressed as

$M=\frac{m}{n}$

Here, $M$ is the molar mass, $m$ is the given mass and $n$ is the number of moles.

The mole fraction of compound can be calculated as

${\chi }_A=\frac{n_A}{n_A+n_B}$

Answer to Problem 139AP

Solution: $2.05\times {10}^{-5}\text{ mmHg}$, $8.90\times {10}^{-5}{}^{\circ }\text{C}$, $2.50\times {10}^{-5}{}^{\circ }\text{C}$ and $0.889\text{ mmHg}$.

Explanation of Solution

Given information: Molar mass of the enzyme is $23930\text{ g/mol}$.

Mass of enzyme is $0.100\text{ g}$.

Mass of water is $150\text{ g}$.

Temperature is ${25}^{\circ }\text{C or }298\text{ K}$.

Vapor pressure of pure water is $23.76\text{ mmHg}$.

Calculate the number of moles of lysozyme by dividing the given mass of enzyme with the molar mass of the enzyme. It is calculated as

$\begin{array}{c}{n}_{lysozyme}=\frac{m_{lysozyme}}{M_{lysozyme}}\\ =\frac{0.100\text{ g}}{13930\text{ g/mol}}\\ =7.18\times {10}^{-6}\text{ mol}\end{array}$

Calculate the number of moles of water by dividing the given mass of water with the molar mass of water. It is calculated as

$\begin{array}{c}{n}_{water}=\frac{m_{water}}{M_{water}}\\ =\frac{\text{150 g}}{18.02\text{ g/mol}}\\ =8.32\text{ mol}\end{array}$

Calculate the mole fraction of lysozyme as

$\begin{array}{c}{\chi }_{lysozyme}=\frac{n_{lysozyme}}{n_{lysozyme}+n_{water}}\\ =\frac{7.18\times {10}^{-6}\text{ mol}}{\left(7.18\times {10}^{-6}\text{ mol}\right)+8.32\text{ mol}}\\ =8.63\times {10}^{-7}\end{array}$

The lowering in vapor pressure is calculated as follows by using the expression given below:

$\begin{array}{l}\Delta P={\chi }_{lysozyme}{P}^{\circ }{}_{water}\\ \end{array}$

Substitute the values of mole fraction and vapor pressure of water

$\begin{array}{c}\Delta P=\left(8.63\times {10}^{-7}\right)\left(23.76\text{ mmHg}\right)\\ =2.05\times {10}^{-5}\text{ mmHg}\end{array}$

Freezing point depression is calculated by using the expression given below:

$\Delta T=K_f.m$

Here, $K_f$ is the freezing point constant and $m$ is the molality.

Substitute the required values

$\begin{array}{c}\Delta T=\left({1.86}^{\circ }\text{C}/m\right)\left(\frac{7.18\times {10}^{-6}}{0.150\text{ kg}}\right)\\ =8.90\times {10}^{-5}{}^{\circ }\text{C}\end{array}$

Boiling point elevation is calculated by using the expression given below:

$\Delta T=K_b.m$

Here, $K_b$ is the boiling point constant and $m$ is the molality.

Substitute the required values

$\begin{array}{c}\Delta T=\left({0.52}^{\circ }\text{C}/m\right)\left(\frac{7.18\times {10}^{-6}\text{ mol}}{0.150\text{ kg}}\right)\\ =2.50\times {10}^{-5}{}^{\circ }\text{C}\end{array}$

Calculate the osmotic pressure by using the expression given below:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, $M$ is the molarity, $\text{R}$ is the gas constant, and $T$ is the temperature.

Assuming the density of solution is $1.00\text{ g/mL}$, substitute $\frac{7.18\times {10}^{-6}\text{ mol}}{0.150\text{ L}}$ for $M$, $0.0821\text{ L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $\text{R}$, and $298\text{ K}$ for $T$

$\begin{array}{c}\pi =\left(\frac{7.18\times {10}^{-6}\text{ mol}}{0.150\text{ L}}\right)(0.0821L\cdot \text{atm/mol}\cdot K)(298K)\\ =1.17\times {10}^{-3}atm\\ =0.889\text{ mmHg}\end{array}$

Conclusion

The vapor pressure lowering, depression in freezing point, elevation in boiling point, and atmospheric pressure of the solution are $2.05\times {10}^{-5}\text{ mmHg}$, $8.90\times {10}^{-5}°\text{C}$, $2.50\times {10}^{-5}°\text{C}$, and $0.889\text{ mmHg}$, respectively.