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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

Expert verified
The freezing-point depression of the given solution is \(2.06 \times 10^{-5} ^{\circ}C\), and the osmotic pressure is \(2.72 \times 10^{-4} \: \text{atm}\). Comparing these values, osmotic pressure provides a more sensitive measure and is less prone to experimental error, making it a better method for determining the molar masses of large molecules.

Step by step solution

01

Calculate the molality of the solution

First, we need to calculate the molality (m) of the solution, which is the number of moles of solute per kilogram of solvent. To do this, we need to determine the number of moles of the protein (n) and the mass of the solvent (wright_air_playlist). Given that the protein has a concentration of 1.0 g/L and a molar mass of 9.0 x 10^4 g/mol, we can calculate the number of moles as: \( n = \frac{1.0 \text{g}}{9.0 \times 10^{4} \text{g/mol}} = 1.11 \times 10^{-5}~\text{mol} \) Since the density of the solution is 1.0 g/cm³ and the volume of the solution is 1 L (1000 mL), the mass of the solution is: Mass of solution = volume × density = 1000 mL × 1.0 g/mL = 1000 g Assuming that the mass of the solute (protein) is negligible compared to the mass of the solvent, the mass of the solvent is approximately the same as the mass of the solution. Now we can calculate the molality of the solution: \( m = \frac{n}{\text{mass of solvent (in kg)}} = \frac{1.11 \times 10^{-5} \text{mol}}{1.0~\text{kg}} = 1.11 \times 10^{-5} \text{mol/kg} \)
02

Calculate the freezing-point depression

Now that we have the molality of the solution, we can use the formula for calculating the freezing-point depression: ΔTf = Kf × m Where Kf is the cryoscopic constant for water, which is 1.86 °C kg/mol. ΔTf = 1.86 °C × 1.11 × 10^{-5} mol/kg = 2.06 × 10^{-5} °C
03

Calculate the osmotic pressure

To calculate the osmotic pressure, we will use the formula: π = n/V × RT Where π is the osmotic pressure, n is the number of moles of solute, V is the volume of the solution in liters, R is the ideal gas constant, and T is the temperature in Kelvin. We already know the number of moles of solute (n = 1.11 × 10^{-5} mol) and the volume of the solution (V = 1 L). The ideal gas constant (R) is 0.0821 L atm/mol K, and the temperature is given as 25°C, which converted to Kelvin is: T(K) = 25°C + 273.15 = 298.15 K Now we can calculate the osmotic pressure: π = (1.11 × 10^{-5} mol) / (1 L) × 0.0821 L atm/mol K × 298.15 K = 2.72 × 10^{-4} atm
04

Compare the two methods and determine the better one

We have calculated the freezing-point depression to be 2.06 × 10^{-5} °C and the osmotic pressure to be 2.72 × 10^{-4} atm. Now, we need to determine which of these would be more suitable for determining the molar masses of large molecules. Comparing the two values, we can see that the osmotic pressure is an order of magnitude larger than the freezing-point depression. This indicates that osmotic pressure would likely provide a more sensitive measure for determining the molar mass of large molecules since it is less prone to experimental error. Additionally, the osmotic pressure method can be easily adapted to a wide range of solution concentrations, while freezing-point depression can be less accurate for very dilute solutions. So, the osmotic pressure method would be better for determining the molar masses of large molecules.

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Key Concepts

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

Freezing-Point Depression
Freezing-point depression is a colligative property that describes how the addition of a solute to a solvent lowers the temperature at which the solvent freezes. Colligative properties are properties that depend on the number of solute particles in a solution, not on the identity of the solute. The presence of solute particles disrupts the formation of the crystalline structure of ice, requiring a lower temperature to freeze the solvent.

To calculate the freezing-point depression, we use the formula:
\[ \Delta T_f = K_f \times m \]
where \( \Delta T_f \) is the change in freezing temperature, \( K_f \) is the cryoscopic constant for the solvent (which is a measure of how the solvent's freezing point changes as the concentration of the solute increases), and \( m \) is the molality of the solution. Molality is defined as moles of solute per kilogram of solvent. For water, the cryoscopic constant is typically \( 1.86 \, \text{°C kg/mol} \).

For very dilute solutions, such as the protein solution in the exercise, freezing-point depression is often too small to measure accurately, which affects its utility for determining molar masses of large molecules.
Osmotic Pressure
Osmotic pressure is another important colligative property that helps determine the direction in which water will flow through a semipermeable membrane, separating solutions of different concentrations. This pressure arises due to the presence of solute particles that cannot cross the barrier, thus, water moves towards the more concentrated solution to balance the particle concentration on both sides.

We calculate osmotic pressure using the formula:
\[ \pi = \frac{n}{V} \times RT \]
where \( \pi \) stands for osmotic pressure, \( n \) is the number of moles of solute, \( V \) is the volume of the solution, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature in Kelvin. For precise measurements such as determining molar mass, this property is particularly useful because osmotic pressure changes are significant enough to be detected even in very dilute solutions.
Molality
Molality is a concentration term used in chemistry that describes the amount of solute per mass of solvent and is expressed as moles per kilogram. Unlike molarity, which depends on the volume of the solution, molality is independent of temperature and pressure because mass does not change with these conditions. The formula to calculate molality is:
\[ m = \frac{n}{\text{mass of solvent (kg)}} \]
where \( n \) is the number of moles of solute. One of the benefits of using molality in colligative property calculations is its reliability under different environmental conditions. For example, the calculations of freezing-point depression and boiling-point elevation requiring molality can yield accurate results regardless of external temperature or pressure variations.
Cryoscopic Constant
The cryoscopic constant, often denoted by \( K_f \), is a proportionality constant that relates the molality of an ideal solution to the degree of freezing-point depression. It is different for each solvent and reflects how sensitive the solvent's freezing point is to the concentration of the solute. The cryoscopic constant is key to calculating the freezing-point depression, and knowing it allows chemists to determine the molar mass of an unknown solute by measuring the solution's freezing point.

When designing experiments to measure molar mass, it is crucial to understand the magnitude of the cryoscopic constant. For solvents with higher cryoscopic constants, even small amounts of solute can create a measurable change in freezing point, enabling more accurate determinations of molar masses, particularly with large, complex molecules, like proteins.

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Most popular questions from this chapter

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 normality of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{HCl}\) b. \(0.105 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(5.3 \times 10^{-2} M \mathrm{H}_{3} \mathrm{PO}_{4}\) d. \(0.134 \mathrm{M} \mathrm{NaOH}\) e. \(0.00521 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) What is the equivalent mass for each of the acids or bases listed above?

A \(2.00-\mathrm{g}\) sample of a large biomolecule was dissolved in \(15.0 \mathrm{~g}\) carbon tetrachloride. The boiling point of this solution was determined to be \(77.85^{\circ} \mathrm{C}\). Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is \(5.03^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\), and the boiling point of pure carbon tetrachloride is \(76.50^{\circ} \mathrm{C}\).

You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?

Anthraquinone contains only carbon, hydrogen, and oxygen. When \(4.80 \mathrm{mg}\) anthraquinone is burned, \(14.2 \mathrm{mg} \mathrm{CO}_{2}\) and \(1.65 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) are produced. The freezing point of camphor is lowered by \(22.3^{\circ} \mathrm{C}\) when \(1.32 \mathrm{~g}\) anthraquinone is dissolved in \(11.4 \mathrm{~g}\) camphor. Determine the empirical and molecular formulas of anthraquinone.

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