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Chapter 13: Problem 81

Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing \(0.150 \mathrm{~g}\) of this enzyme in \(210 \mathrm{~mL}\) of solution has an osmotic pressure of \(0.127 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\). What is the molar mass of lysozyme?

Short Answer

Expert verified
The molar mass of lysozyme is approximately \(14,019\,\text{g/mol}\).

Step by step solution

01

Write Down the Given Information

We are given the mass of lysozyme as \(0.150\,\text{g}\), the volume of the solution as \(210\,\text{mL}\) (which is \(0.210\,\text{L}\)), the osmotic pressure \(\pi\) as \(0.127\,\text{kPa}\), and the temperature \(T\) as \(25^{\circ}C\) (convert to Kelvin: \(T = 25 + 273.15 = 298.15\,\text{K}\)).
02

Convert Osmotic Pressure to Appropriate Units

Convert the osmotic pressure from kilopascals to atmospheres since the gas constant \(R\) is usually in \(\text{L atm mol}^{-1} \text{K}^{-1}\): \(0.127\,\text{kPa} \times \frac{1\,\text{atm}}{101.325\,\text{kPa}} = 0.001253\,\text{atm}\).
03

Use the Osmotic Pressure Formula

The formula for osmotic pressure is \(\pi = \frac{n}{V}RT\), where \(n\) is the number of moles, \(V\) is the volume in liters, \(R\) is the ideal gas constant (use \(0.0821\,\text{L atm mol}^{-1} \text{K}^{-1}\)), and \(T\) is the temperature in Kelvin. Rearrange the formula to solve for \(n\): \[n = \frac{\pi V}{RT}\]. Substitute the values: \[n = \frac{0.001253 \times 0.210}{0.0821 \times 298.15} = 1.07 \times 10^{-5} \text{ mol}\].
04

Calculate the Molar Mass

Molar mass \(M\) is calculated by dividing the mass of the substrate (\(0.150\,\text{g}\)) by the number of moles calculated in the previous step:\[M = \frac{0.150}{1.07 \times 10^{-5}} = 14018.69\,\text{g/mol}\]. Thus, the molar mass of lysozyme is approximately \(14,019\,\text{g/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Enzymes
Enzymes are fascinating proteins that act as catalysts in various biological processes. They speed up reactions without being consumed in the process. Lysozyme, for instance, breaks down the cell walls of bacteria, helping to protect organisms from bacterial infections.

Here鈥檚 what makes enzymes unique and vital in biological contexts:
  • **Specificity**: Each enzyme caters to a particular substrate. Lysozyme specifically targets bacterial cell walls.
  • **Efficiency**: Enzymes can accelerate chemical reactions by factors of millions.
  • **Regulation**: Enzymes can be turned on or off, offering cells precise control over metabolic pathways.
Without enzymes, many reactions necessary for life would proceed too slowly to sustain living organisms. Understanding these proteins helps us grasp how biological systems operate at the molecular level.
Molar Mass Calculation
Calculating molar mass is crucial in chemistry, especially when dealing with reactions and substances. By understanding the concept of molar mass, you can effectively determine how much of a substance is involved in a given chemical process. To calculate molar mass:
  • First, identify the mass of the substance. In our example, the mass of lysozyme is given as 0.150 g.
  • Next, determine the number of moles of the substance. Here, we used the osmotic pressure formula to find the moles, resulting in \(1.07 \times 10^{-5}\) mol.
  • Finally, divide the mass by the number of moles to find the molar mass. This approach revealed lysozyme鈥檚 molar mass to be approximately 14019 g/mol.
This process underscores how we can use physical characteristics like osmotic pressure to derive essential quantitative information about a compound. Understanding molar mass is vital for various applications, from stoichiometry to molecular biology.
The Ideal Gas Constant and Its Role
The ideal gas constant, denoted as \(R\), is a fundamental part of calculations involving gases. It appears in various equations, including the ideal gas law and osmotic pressure equation, which we used in the lysozyme example.
In the context of the osmotic pressure formula \(\pi = \frac{n}{V}RT\), \(R\) = 0.0821 L atm mol鈦宦� K鈦宦� is the ideal gas constant that relates pressure, volume, and temperature of gases. Here鈥檚 why \(R\) is significant:
  • **Unifying Factor**: \(R\) provides a bridge between molar volume and gas behavior under various conditions.
  • **Consistency**: The constant's value ensures that calculations yield consistent and comparable results.
  • **Versatility**: Apart from gases, \(R\) is useful in scenarios like calculating osmotic pressures of solutions, as we did with lysozyme.
Understanding \(R\) and its utility in physics and chemistry helps us explain and predict the behavior of substances in different states, making it a cornerstone concept in science.

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions: (a) \(1.50 \mathrm{~L}\) of \(0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) \(225 \mathrm{~g}\) of a solution that is \(0.65 \mathrm{~m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), starting with the solid solute; (c) 1.20 \(\mathrm{L}\) of a solution that is \(15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 M\) solution of HCl that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with \(6.0 \mathrm{MHCl}\).

Aerosols are important components of the atmosphere. Does the presence of aerosols in the atmosphere increase or decrease the amount of sunlight that arrives at the Earth's surface, compared to an "aerosol-free" atmosphere? Explain your reasoning.

Commercial concentrated aqueous ammonia is \(28 \% \mathrm{NH}_{3}\) by mass and has a density of \(0.90 \mathrm{~g} / \mathrm{mL}\). What is the molarity of this solution?

A lithium salt used in lubricating grease has the formula \(\mathrm{LiC}_{n} \mathrm{H}_{2 n+1} \mathrm{O}_{2}\). The salt is soluble in water to the extent of \(0.036 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). The osmotic pressure of this solution is found to be \(7.61 \mathrm{kPa}\). Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of \(n\) in the formula for the salt.

The Baltic Sea has a salinity of \(1.0 \%\), that is, its water contains \(10 \mathrm{~g}\) of salt for every liter of solution. Assuming that the solute consists entirely of \(\mathrm{NaCl}\) (in fact, over \(90 \%\) of the salt is indeed \(\mathrm{NaCl}\), calculate the osmotic pressure of this seawater at \(15^{\circ} \mathrm{C}\).
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