Question

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing 1.0 \(\mathrm{g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{~g} / \mathrm{mol}\) ) if the density of the solution is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\). b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

Short answer

The freezing-point depression of the aqueous protein solution is \(2.06 \times 10^{-5}\) °C, and the osmotic pressure is \(2.73 \times 10^{-3}\) atm. Osmotic pressure would be a better choice for determining the molar masses of large molecules because it has a higher value, making it easier to measure in experiments.

Step-by-step solution

Calculate molality of the protein solution

Given that there is 1.0 g/L of protein, we convert this value to moles by dividing by the protein's molar mass: moles of protein = 1.0 g / (9.0 × 10^4 g/mol) = \(1.11 \times 10^{-5} \mathrm{mol/L}\) Now, let's convert the protein concentration from mol/L to mol/kg solvent. Since the density of the solution is 1.0 g/cm³, we will use the following conversion factors: 1 L = 1000 cm³ 1 g/cm³ = 1 kg/L So, the concentration of protein = \(1.11 \times 10^{-5} \mathrm{mol/L} \times \frac{1\;\mathrm{kg}}{1\;\mathrm{L}} = 1.11 \times 10^{-5} \mathrm{mol/kg}\) The molality of protein in the solution is \(1.11 \times 10^{-5}\) mol/kg.

Calculate freezing-point depression

To calculate the freezing-point depression (ΔTf), we use the formula ΔTf = Kf * molality. For water, Kf = 1.86 °C·kg/mol. Therefore: ΔTf = (1.86 °C·kg/mol) * (1.11 × 10^{-5} mol/kg) = \(2.06 \times 10^{-5}\) °C

Calculate osmotic pressure

To calculate the osmotic pressure (Π), we use the formula Π = n/V * RT. Here, we are given the temperature (T) as 25°C. First, let's convert the temperature to Kelvin: T = 25°C + 273.15 K = 298.15 K R = 0.0821 L·atm/mol·K (ideal gas constant) Now, we can calculate the osmotic pressure: Π = (1.11 × 10^{-5} mol/L) * (0.0821 L·atm/mol·K) * (298.15 K) = \(2.73 \times 10^{-3}\) atm

Determine which colligative property is better for determining molar masses

Comparing the values obtained for freezing-point depression and osmotic pressure, the freezing-point depression is considerably smaller (\(2.06 \times 10^{-5}\) °C) than the osmotic pressure (\(2.73 \times 10^{-3}\) atm). Therefore, it would be more challenging to accurately measure the freezing-point depression, especially when dealing with large molecules. Thus, osmotic pressure would be a better choice for determining the molar masses of large molecules, as its value is higher, making it easier to measure in experiments.

Key concepts

Freezing-Point Depression

Understanding freezing-point depression can be quite intriguing! It's a colligative property, which means it depends on the amount of solute particles in a solution, not their identity. Imagine a pure solvent—like water; it freezes at a specific temperature, right? Now, throw in some solute, and things change. The solute particles interrupt the forming of the ice crystal lattice, making it harder for the solvent to solidify, so the freezing point goes down.

This phenomenon has a formula: \( \Delta T_f = K_f \times m \) where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the freezing-point depression constant specific to the solvent, and \( m \) the molality of the solution. Remember, the molality (\( m \) is moles of solute per kg of solvent. Through this relationship, we see why a higher solute concentration can cause a greater freezing-point drop—it's all about how these solute particles interfere!

Osmotic Pressure

Now, let's dive into osmotic pressure, another intriguing colligative property. Osmosis is like a survival instinct for solutions—they want to balance solute concentration on both sides of a semi-permeable membrane. Osmotic pressure is the force required to stop this natural flow.

To work out osmotic pressure, we use the formula \( \Pi = \frac{n}{V} \times RT \), where \( \Pi \) is the osmotic pressure, \( n \) is the number of moles of solute, \( V \) is the volume of the solution, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. Basically, as you increase the solute amount (\( n \) in moles), the osmotic pressure shoots up because there are more particles that want to move across the membrane.

Molar Mass

Cracking the Code of Molar Mass

Picture molar mass as a signature—each substance has its own. Technically, it's the mass of one mole of a substance. But what's a mole? It's a gigantic number of particles (Avogadro's number, to be precise), which is similar to saying a 'dozen' stands for '12'.

By knowing the molar mass, chemists can measure out exact amounts of a substance. It's essentially the bridge between the microscopic world of atoms and molecules and the macroscopic world we live in. So when scientists talk about grams per mole (\( g/mol \)), think of it as a recipe that ensures you're using just the right amount of each ingredient.

Molality

What's the Scoop on Molality?

Molality might sound like molarity, but it's a bit different. While molarity is moles of solute per liter of solution, molality is moles of solute per kilogram of solvent. Why is this important? Because molality isn't affected by temperature! It keeps the calculation consistent, even if the solution expands or contracts with heat, something that can't be said for molarity.

Mathematically, molality (\( m \) is: \( m = \frac{moles\;of\;solute}{kilograms\;of\;solvent} \). It's ideal for studying temperature-dependent processes like boiling point elevation or, you guessed it, freezing-point depression. A high molality means you've got a strong-willed solution that's reluctant to freeze—just like a determined person refusing to stop despite the cold!