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An aqueous solution of \(10.00 \mathrm{~g}\) of catalase, an enzyme found in the liver, has a volume of \(1.00 \mathrm{~L}\) at \(27^{\circ} \mathrm{C}\). The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be \(0.745\) torr. Calculate the molar mass of catalase.

Short Answer

Expert verified
The molar mass of catalase is approximately \(250,500 \mathrm{~g/mol}\).

Step by step solution

01

Convert temperature to Kelvin

First, convert the temperature from Celsius to Kelvin using the formula: K = °C + 273.15 T = 27°C + 273.15 = 300.15 K
02

Convert osmotic pressure to atm

Next, convert the osmotic pressure from torr to atm using the conversion factor: 1 atm ≈ 760 torr Π = 0.745 torr × (1 atm / 760 torr) ≈ 0.00098026 atm
03

Rewrite osmotic pressure formula

We can rewrite the osmotic pressure formula to find the number of moles (n) of solute: n = ΠV/RT
04

Find the number of moles of solute

Now, we can substitute the known values into the formula to find the number of moles of catalase: n = (0.00098026 atm)(1.00 L) / (0.08206 L·atm/mol·K)(300.15 K) ≈ 3.996 × 10⁻⁵ mol
05

Calculate molar mass of catalase

To find the molar mass of catalase, divide the mass of catalase by the number of moles: Molar mass = (10.00 g) / (3.996 × 10⁻⁵ mol) ≈ 250,500 g/mol So, the molar mass of catalase is approximately 250,500 g/mol.

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Key Concepts

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

Aqueous Solution
In chemistry, an aqueous solution is a solution where water is the solvent. The term "aqueous" comes from the Latin word for water, "aqua." When a substance is dissolved in water, it is said to be in an aqueous solution. A common example is table salt (sodium chloride) dissolved in water.

In the case of our exercise, the enzyme catalase is dissolved to create an aqueous solution. One primary characteristic of aqueous solutions is that they often facilitate the movement of ions and molecules, enabling chemical reactions. This characteristic is crucial for biological processes like enzyme activity. Understanding aqueous solutions helps in predicting how substances behave when mixed with water, including changes in solubility, conductivity, and reaction rates.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is a key concept in chemistry that helps quantify the amount of a substance. To find the molar mass of a compound, sum up the atomic masses of all the elements present in one mole of that compound.

In the given problem, we calculated the molar mass of the enzyme catalase by dividing the mass of the substance (10.00 g) by the number of moles calculated using the osmotic pressure formula. This calculation revealed that the molar mass of catalase is approximately 250,500 g/mol. Calculating molar mass is essential in determining how substances interact on a molecular level.
Catalase Enzyme
Catalase is a biochemical enzyme found in nearly all living organisms exposed to oxygen. Its primary function is to catalyze the decomposition of hydrogen peroxide into water and oxygen, a critical process for protecting cells from oxidative damage. As a protein, catalase consists of chains of amino acids and other complex structures.

This enzyme has a strikingly high molar mass, as demonstrated in the exercise, with approximately 250,500 g/mol. Understanding enzymes like catalase is essential in biochemistry because they play a role in various metabolic processes, helping organisms manage chemical reactions necessary for life. The study of such enzymes is crucial for advancements in medicine and biotechnology.
Conversion of Units
The conversion of units is a fundamental skill in chemistry and science in general. It involves changing a quantity expressed in one type of unit to another through multiplication or division. Common conversions include converting units of pressure, temperature, volume, and mass to their respective equivalents in other systems.

In the exercise, we converted the temperature from Celsius to Kelvin using the formula \( K = °C + 273.15 \), and pressure from torr to atm using the conversion factor \( 1 \text{ atm} \approx 760 \text{ torr} \). These conversions allow for the proper use of physical laws and formulas, such as the ideal gas law in this context. Mastery of unit conversion is crucial for accurately interpreting scientific data and solving equations effectively.

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Most popular questions from this chapter

Calculate the solubility of \(\mathrm{O}_{2}\) in water a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration \((\mathrm{mol} / \mathrm{L})\).

An aqueous solution containing \(0.250\) mole of \(\mathrm{Q}\), a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{~g}\) water freezes at \(-2.79^{\circ} \mathrm{C}\). What is the van't Hoff factor for Q? The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\). What is the formula of \(\mathrm{Q}\) if it is \(38.68 \%\) chlorine by mass and there are twice as many anions as cations in one formula unit of \(\mathrm{Q}\) ?

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C}\). c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

Calculate the freezing point and boiling point of an antifreeze solution that is \(50.0 \%\) by mass of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) in water. Ethylene glycol is a nonelectrolyte.

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner \(\mathrm{R} . \mathrm{B}\). Woodward. It is used as a tranquilizer and sedative. When \(1.00 \mathrm{~g}\) reserpine is dissolved in \(25.0 \mathrm{~g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\) for camphor is \(40 .{ }^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) ). Calculate the molality of the solution and the molar mass of reserpine.

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