Question

Lysozyme is an enzyme that cleaves bacterial cell walls. A sample of lysozyme extracted from egg white has a molar mass of $13,930\mathrm{g}$. A quantity of $0.100\mathrm{g}$ of this enzyme is dissolved in $150\mathrm{g}$ of water at ${25}^{\circ }\mathrm{C}$. 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 ${25}^{\circ }\mathrm{C}$ is $23.76\mathrm{mmHg}.)$

Short answer

Vapor-pressure lowering: 0.00017 mmHg; Freezing point depression: 0.0000891 °C; Boiling point elevation: 0.0000245 °C; Osmotic pressure: 0.89 mmHg.

Step-by-step solution

Calculate mole fraction

First, determine the amount of moles of lysozyme. Use the formula $n=\frac{m}{M}$, where $m=0.100\text{ g}$ is the mass of the solute, and $M=13930\text{ g/mol}$ is the molar mass.$$n=\frac{0.100}{13930}=7.18\times {10}^{-6}\text{ mol}$$The moles of water can be calculated from its mass: $150\text{ g}$ of water is equivalent to $\frac{150}{18}=8.33\text{ mol of water}$. Now, find the mole fraction $\chi$ of water:$${\chi }_{\text{water}}=\frac{8.33}{8.33+7.18\times {10}^{-6}}\approx 0.999999$$

Calculate vapor-pressure lowering

The change in vapor pressure can be expressed as:$$\mathrm{\Delta }P=P_0-P=P_0\cdot (1-{\chi }_{\text{water}})\approx P_0\cdot n_{\text{lysozyme}}$$For very small solute amounts, approximately:$$\mathrm{\Delta }P\approx 23.76\times 7.18\times {10}^{-6}=1.70\times {10}^{-4}\text{ mmHg}$$

Calculate the depression in freezing point

Utilize the formula for freezing point depression $\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m$. Here, $i=1$ for a non-electrolyte, $K_f=1.86\text{ °C·kg/mol}$, and compute $m$, the molality:$$m=\frac{7.18\times {10}^{-6}}{0.150}=4.79\times {10}^{-5}\text{ mol/kg}$$$$\mathrm{\Delta }{T}_f=1\cdot 1.86\cdot 4.79\times {10}^{-5}=8.91\times {10}^{-5}\text{ °C}$$

Calculate the elevation in boiling point

Use the boiling point elevation formula $\mathrm{\Delta }{T}_b=i\cdot K_b\cdot m$. Here, $i=1$ and $K_b=0.512\text{ °C·kg/mol}$:$$\mathrm{\Delta }{T}_b=1\cdot 0.512\cdot 4.79\times {10}^{-5}=2.45\times {10}^{-5}\text{ °C}$$

Calculate the osmotic pressure

Apply the osmotic pressure equation $\mathrm{\Pi }=nRT/V$, or $\mathrm{\Pi }=cRT$ with $c$ as molarity. Water volume $V$ can be approximated as $0.150\text{ L}$:Molarity $c$ is:$$c=\frac{7.18\times {10}^{-6}}{0.150}=4.79\times {10}^{-5}\text{ mol/L}$$At ${25}^{ext°}C$, $R=0.08206\text{ L atm/mol K}$ and $T=298\text{ K}$:$$\mathrm{\Pi }=4.79\times {10}^{-5}\times 0.08206\times 298=1.17\times {10}^{-3}\text{ atm}\approx 0.89\text{ mmHg}$$

Key concepts

Vapor Pressure Lowering

When a non-volatile solute like lysozyme is dissolved in water, it causes the vapor pressure of the solution to be lower than that of pure water. This is known as vapor pressure lowering. When you dissolve a solute, it takes up space at the surface of the liquid, where molecules escape to form vapor. As a consequence, fewer water molecules are at the surface to escape into the vapor phase, leading to a lower vapor pressure.
To calculate this drop in vapor pressure, we can use Raoult's Law, which states that the reduction in vapor pressure is proportional to the mole fraction of the solute. The equation is:

• $$\mathrm{\Delta }P=P_0\cdot n_{\text{solute}}$$

Where $P_0$ is the vapor pressure of the pure solvent. This concept is crucial in understanding why solutions have different physical properties than their pure components.

Freezing Point Depression

Freezing point depression is a colligative property that describes the lowering of the freezing point when a solute is dissolved in a solvent. When lysozyme is added to water, solute particles interfere with the formation of the solid crystal lattice necessary for the solvent to freeze. This means the solution's freezing point will be lower than that of the pure solvent.
The magnitude of the freezing point depression can be calculated using:

• $$\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m$$

Where: - $i$ is the van 't Hoff factor (for lysozyme, $i=1$ since it does not ionize), - $K_f$ is the cryoscopic constant of the solvent, and - $m$ is the molality of the solution.
This property is especially useful in areas like antifreeze in car radiators, where added solutes prevent the coolant from freezing in cold temperatures.

Boiling Point Elevation

Similar to freezing point depression, boiling point elevation is another colligative property experienced by a solution. It refers to the increase in boiling point brought about when a solute such as lysozyme is added to a solvent like water. Solute particles cause a decrease in the solvent's vapor pressure, requiring a higher temperature to reach the boiling point.
To find how much the boiling point increases, we use:

• $$\mathrm{\Delta }{T}_b=i\cdot K_b\cdot m$$

In this formula:- $i$ remains the van 't Hoff factor,- $K_b$ is the ebullioscopic constant for the solvent, and - $m$ is the molality.
Understanding boiling point elevation allows chemists and engineers to design processes involving temperature-sensitive materials.

Osmotic Pressure

Osmotic pressure is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane. It is a vital property in biological and chemical processes because it helps maintain proper cell function.
When lysozyme is dissolved in water, the osmotic pressure can be calculated using the formula:

• $$\mathrm{\Pi }=cRT$$

Where:- $c$ is the molarity of the solution, - $R$ is the ideal gas constant, and - $T$ is the temperature in Kelvins.
Osmotic pressure is crucial in settings ranging from the dialysis process to preserving food. It also helps create concentration gradients within cells, essential for nutrient uptake and waste expulsion.