Question

A 0.15 -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

The molar mass of the protein is approximately 74,759 g/mol.

Step-by-step solution

Convert temperature to Kelvin and pressure to atm

First, we need to convert the temperature from Celsius to Kelvin and the pressure from torr to atm. Temperature in Kelvin (T) = 25°C + 273.15 = 298.15 K 1 atm = 760 torr Osmotic pressure in atm (π) = 18.6 torr * (1 atm / 760 torr) = 0.02447 atm

Rearrange the osmotic pressure equation to find the number of moles and compute n

Next, we rearrange the osmotic pressure equation to find the number of moles (n): n = π * V / (R * T) Use the given volume (V = 2.0 mL) and convert it to liters, knowing there are 1000 mL in one liter: V = 2.0 mL * (1 L / 1000 mL) = 0.002 L Now, substitute the values of π, V, R, and T into the equation: n = (0.02447 atm) * (0.002 L) / (0.0821 L atm K⁻¹ mol⁻¹) * (298.15 K) n ≈ 2.007 * 10⁻⁶ moles

Calculate the molar mass of the protein

Finally, use the mass of the protein and the number of moles to determine the molar mass (M): M = mass / n The mass of the protein is given as 0.15 g. Use this value and the calculated number of moles (n ≈ 2.007 * 10⁻⁶) to find the molar mass: M ≈ (0.15 g) / (2.007 * 10⁻⁶ moles) M ≈ 74758.72 g/mol The molar mass of the protein is approximately 74,759 g/mol.

Key concepts

Molar Mass

Understanding the molar mass of a substance is key in chemistry, as it relates to the mass of one mole (approximately 6.022\( \times \)10\(^{23}\) entities) of that substance. The molar mass is expressed in grams per mole (g/mol) and is calculated by dividing the mass of a sample by the number of moles present.

In the context of the osmotic pressure calculation, the molar mass becomes particularly important. As the steps provided show, after determining the number of moles from the osmotic pressure using the equation \(\pi = \frac{nRT}{V}\), the molar mass can be found with \(M = \frac{mass}{n}\). Here, mass is the amount of substance in grams and n is the number of moles calculated. Thus, accurate measurement of the sample mass and careful calculation of the moles are crucial to determining the correct molar mass of a substance, such as the protein in the exercise.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules and not on the nature of the chemical species present. They include freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.

Osmotic pressure, which is the focus in the exercise, is a colligative property that measures the tendency of a solution to take in water by osmosis. It's influenced by soluble particles and can be harnessed to calculate the molar mass of unknown substances, such as proteins. Knowing how to harness these properties can provide important insights into solution chemistry and aid in the determination of substance characteristics, including molar mass, by the mere assessment of the behavior of a solution.

Gas Laws

Gas laws help us understand the relationship between pressure, volume, temperature, and the number of moles of a gas. In the equation used to calculate osmotic pressure, \(\pi = \frac{nRT}{V}\), the gas constant (R) and the temperature in Kelvins (T) directly relate to the ideal gas law, which is given by \(PV = nRT\), where P is the pressure, V is the volume, and n is the amount of gas in moles.

In the solution steps, we see this relationship applied analogously to a solution, as osmotic pressure is conceptually similar to the pressure exerted by a gas. This resemblance allows the use of the gas law constants to calculate the properties related to solutions, such as determining the number of moles of solute in a given volume, which then feeds into the calculation of molar mass.

Solution Chemistry

Solution chemistry involves studying the physical and chemical interactions that occur when a solute is dissolved in a solvent. It encompasses understanding how these interactions affect the properties of the resulting solution, leading to practical applications like the calculation of osmotic pressure.

Osmotic pressure, a fundamental concept in solution chemistry, requires careful preparation of the solution and precise measurements. As shown in the solution steps, knowing the volume of the solution and its temperature allows for the calculation of osmotic pressure. This can then be used with other properties, such as the molar mass, to understand more deeply the behavior of the solute within the solvent. Mastery of solution chemistry concepts is essential for any chemist, as it can lead to discoveries about solubility, reactivity, and the colligative properties of substances.