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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 \(13.89 \frac{\mathrm{g}}{\mathrm{mol}}\).

Step by step solution

01

Write down the given information

We have the following information: - Mass of lysozyme: \(0.150 \mathrm{~g}\) - Volume of the solution: \(210 \mathrm{~mL}\) - Osmotic pressure: \(0.127 \mathrm{kPa}\) - Temperature: \(25^{\circ} \mathrm{C}\)
02

Convert the given information to appropriate units

We need to convert the temperature to Kelvin and the volume of the solution to liters: - Temperature in Kelvin: \(25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) - Volume of the solution in liters: \(\frac{210 \mathrm{~mL}}{1000} = 0.210 \mathrm{~L}\)
03

Use the osmotic pressure equation to solve for molar concentration

The osmotic pressure equation is given by: \(Π = \frac{n}{V} \times R \times T\) Where: - \(Π\) is the osmotic pressure - \(n\) is the number of moles of solute - \(V\) is the volume of the solution - \(R\) is the ideal gas constant (\(R = 8.314 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}\)) - \(T\) is the temperature in Kelvin We can solve for the number of moles, \(n\), by re-arranging the equation: \(n = \frac{Π \times V}{R \times T}\) Now plug in the given values: \(n = \frac{0.127 \times 10^3 \times 0.210}{8.314 \times 298.15}\) \(n \approx 0.01080 \mathrm{~mol}\)
04

Calculate the molar mass

Now that we have the number of moles, we can calculate the molar mass by dividing the mass of lysozyme by the moles: Molar mass = \(\frac{mass}{moles}\) Molar mass = \(\frac{0.150 \mathrm{~g}}{0.01080 \mathrm{~mol}}\) Molar mass \(\approx 13.89 \frac{\mathrm{g}}{\mathrm{mol}}\) The molar mass of lysozyme is approximately \(13.89 \frac{\mathrm{g}}{\mathrm{mol}}\).

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Key Concepts

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

Molar Mass Calculation
Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance. It is usually expressed in grams per mole (g/mol). Understanding molar mass is crucial when dealing with chemical reactions, solutions, and calculations involving molarity.
To find the molar mass of a compound, you need the total mass of the substance and the number of moles present. The relationship is given by the formula:
  • Molar Mass = \(\frac{\text{mass (g)}}{\text{number of moles (mol)}}\)
In the context of our problem, we used the mass of lysozyme and the number of moles (determined through osmotic pressure calculations) to find its molar mass. This approach is not only applicable to lysozyme but also to a wide range of substances used in chemistry.
Remember, accurate conversions and unit management are important when calculating molar mass, as small errors can significantly affect the result.
Enzyme Lysozyme
Lysozyme is an enzyme renowned for its ability to break down bacterial cell walls. This biological catalyst is particularly effective against certain bacteria by cleaving the bonds in the cell wall polysaccharides.
Lysozyme plays a significant role in the innate immune system, providing protection against bacterial infections. Found naturally in many bodily fluids like tears and saliva, it functions as a first line of defense.
Its enzymatic activity has made lysozyme a subject of various biochemical studies, particularly in understanding its structure and mechanism of action. Additionally, lysozyme is often used in experimental settings to study enzyme-related phenomena, including reactions and enzyme-substrate interactions.
Ideal Gas Constant
The Ideal Gas Constant, denoted as \(R\), is a crucial component of the ideal gas law equation. It embodies the proportionality constant linking pressure, volume, temperature, and moles of a gas. Its value is typically \(8.314 \frac{\text{J}}{\text{mol} \cdot \text{K}}\), and it allows for the conversion of these properties into consistent, meaningful results during calculations.
In the context of osmotic pressure situations, the ideal gas constant can be used beyond gases alone. This relation is significant in solutions because it helps determine molar concentrations and can be extended to other states of matter.
  • Osmotic Pressure Equation: \(Π = \frac{n}{V} \times R \times T\)
  • \(Π\) represents osmotic pressure.
  • \(n\) is the number of moles.
  • \(V\) is the volume in liters.
  • \(T\) is the temperature in Kelvin.
Understanding and employing the value of \(R\) accurately is essential in performing calculations, such as determining the molar mass, as seen in this exercise. It ensures the precision and validity of the results obtained.

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Most popular questions from this chapter

Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of \(5.00 \mathrm{~g}\) of lauryl alcohol in \(0.100 \mathrm{~kg}\) of benzene freezes at \(4.1^{\circ} \mathrm{C}\). What is the molar mass of lauryl alcohol from this data?

(a) What is the molality of a solution formed by dissolving 1.12 mol of KCl in 16.0 mol of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?

You take a sample of water that is at room temperature and in contact with air and put it under a vacuum. Right away, you see bubbles leave the water, but after a little while, the bubbles stop. As you keep applying the vacuum, more bubbles appear. A friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, causing the water to boil. Another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen, and so forth) that were dissolved in the water. Which friend is mostly likely to be correct? What, then, is responsible for the second batch of bubbles? [Section 13.4]

During a person's typical breathing cycle, the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of \(4.6 \%\) by volume. (a) Calculate the partial pressure of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming \(101.3 \mathrm{kPa}\) pressure and a body temperature of \(37^{\circ} \mathrm{C}\). (b) What is the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming a body temperature of \(37^{\circ} \mathrm{C} ?\)

Consider two solutions, one formed by adding \(150 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to \(1 \mathrm{~L}\) of water and the other formed by adding \(150 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(1 \mathrm{~L}\) of water. Calculate the vapor pressure for each solution at \(25^{\circ} \mathrm{C} ;\) the vapor pressure of pure water at this temperature is \(3.17 \mathrm{kPa}\).

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