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A \(0.15-\mathrm{g}\) sample of a purified protein is dissolved in water to give \(2.0 \mathrm{~mL}\) of solution. The osmotic pressure is found to be \(18.6\) torr at \(25^{\circ} \mathrm{C}\). Calculate the protein's molar mass.

Short Answer

Expert verified
The molar mass of the purified protein is \(3020 \thinspace \text{g/mol}\).

Step by step solution

01

Convert the given values to appropriate units

Convert the temperature from Celsius to Kelvin and the volume from mL to Liters. Given temperature, T = 25°C To convert to Kelvin, add 273 to the Celsius temperature: T = 25°C + 273 = 298 K Given volume, V = 2.0 mL To convert to Liters, divide by 1000: V = 2.0 mL / 1000 = 0.002 L
02

Rearrange the osmotic pressure formula to solve for the number of moles

We will rearrange the osmotic pressure formula to calculate the number of moles (n). n = (Π * V) / (R * T)
03

Substitute the given values and find the number of moles

Substitute the given values (osmotic pressure, volume, temperature, and gas constant) into the rearranged formula to find the number of moles. Given osmotic pressure, Π = 18.6 torr To convert to atm, divide by 760: Π = 18.6 torr / 760 = 0.02447 atm Gas constant, R = 0.08206 L atm/K mol Now substitute the values: n = (0.02447 atm * 0.002 L) / (0.08206 L atm/K mol * 298 K) n = 0.0000496
04

Calculate the molar mass of the protein

Use the mass and the number of moles calculated in Step 3 to find the molar mass of the protein. Molar mass = (mass of the sample) / (number of moles) Given mass, m = 0.15 g Number of moles, n = 0.0000496 Molar mass = 0.15 g / 0.0000496 = 3020 g/mol The molar mass of the purified protein is 3020 g/mol.

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Key Concepts

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

Osmotic Pressure
Understanding osmotic pressure is crucial for calculating the molar mass of substances, such as proteins, in solution. Osmotic pressure (Π) is a colligative property, meaning it depends on the number of solute particles in a solution, not their identity. It's the pressure required to prevent the flow of a pure solvent into a solution through a semipermeable membrane.

When a solute, like a protein, is present in water, it generates osmotic pressure, which can be measured empirically. In the context of the exercise, the protein's osmotic pressure was found to be 18.6 torr, a pressure unit often used in chemistry. This osmotic pressure signifies the tendency of water to move into the solution to balance solute concentrations on both sides of the membrane. By understanding osmotic pressure, we can begin to unravel the properties of the protein solution, ultimately allowing the calculation of the protein's molar mass.
Unit Conversion
Precise unit conversion is an essential step in the realm of scientific calculations to ensure accuracy and uniformity. When we're dealing with osmotic pressure, temperature, and volume, it's important that all measurements are in compatible units before using them in formulas.

For instance, in the given problem, the temperature must be converted from Celsius to Kelvin, a standard unit for thermodynamic equations, by adding 273. Similarly, volume should be in liters (L) for use with the gas constant (R in L atm/K mol). Osmotic pressure itself is often measured in units such as torr; however, for calculations using the ideal gas law, it should be converted to atmospheres (atm) by dividing by 760. This step is crucial to avoid any discrepancies in the calculations leading to the determination of molar mass.
Molar Mass
The molar mass is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). It links the microscale of individual particles to the macroscale that we can measure in the lab. To calculate the molar mass of the protein in the exercise, we first determined the number of moles of the protein dissolved in the solution and then used the mass of the protein sample.

Keeping in mind that a mole is Avogadro's number of particles (approximately 6.02 x 10^23 particles), we then relate the small mass of our protein sample to the colossal number of molecules it contains. By calculating the molar mass, we can deliver extremely useful information regarding the protein, which can be applied to various fields, including pharmaceuticals, bioengineering, and nutrition science.
Colligative Properties
Colligative properties are fundamental characteristics of solutions that depend on the ratio of solute particles to solvent molecules, regardless of the type of solute. These properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Osmotic pressure is one such property we're exploring, especially pertaining to its role in the determination of molar mass, as presented in the exercise. Because these properties are directly proportional to the concentration of solute particles, they can reveal a lot about the solution without needing to identify the solute. Thus, by studying colligative properties like osmotic pressure, we gain insights into molecular quantities and interactions that would otherwise require more complex methods to ascertain.

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Most popular questions from this chapter

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C}\) ? a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}_{3} \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol [ 143 torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water. \()\)

From the following: pure water solution of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(m=0.01)\) in water solution of \(\mathrm{NaCl}(m=0.01)\) in water solution of \(\mathrm{CaCl}_{2}(m=0.01)\) in water Choose the one with the a. highest freezing point. b. lowest freezing point. c. highest boiling point. d. lowest boiling point. e. highest osmotic pressure.

You drop an ice cube (made from pure water) into a saltwater solution at \(0^{\circ} \mathrm{C}\). Explain what happens and why.

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

In a coffee-cup calorimeter, \(1.60 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) was mixed with \(75.0 \mathrm{~g}\) water at an initial temperature \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\). a. Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\text {soln }}\right)\) for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol}\). b. If the enthalpy of hydration for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(-630 . \mathrm{kJ} / \mathrm{mol}\), calculate the lattice energy of \(\mathrm{NH}_{4} \mathrm{NO}_{3} .\)

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