Question

Consider two aqueous solutions, one of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and the other of nitric acid \(\left(\mathrm{HNO}_{3}\right) .\) Both solutions freeze at \(-1.5^{\circ} \mathrm{C}\). What other properties do these solutions have in common?

Short answer

Both solutions share the same colligative properties such as freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.

Step-by-step solution

Understanding Freezing Point Depression

Both solutions freeze at the same temperature of -1.5 °C, which indicates that they have undergone freezing point depression. Freezing point depression is a colligative property that depends on the number of solute particles in the solution, not the identity of the solute.

Determine Molality Using Freezing Point Depression Formula

The freezing point depression (\(\Delta T_f\)) can be used to find the molality (\(m\)) of the solutions through the formula \[\Delta T_f = i \cdot K_f \cdot m\], where \(\Delta T_f = 1.5\,\mathrm{C}\), and \(K_f\) is the cryoscopic constant for water (1.86 °C kg/mol). For both solutions: \[1.5 = i \cdot 1.86 \cdot m\].

Determine the Van't Hoff Factor

For sucrose, \(i = 1\) since it is a non-electrolyte and does not dissociate. For nitric acid, \(i \approx 2\) because it is a strong acid that dissociates completely into \(\mathrm{H}^+\) and \(\mathrm{NO}_3^-\).

Calculate Molality

For the sucrose solution, \(1.5 = 1 \cdot 1.86 \cdot m\) gives \(m \approx 0.806 \ \mathrm{mol/kg}\). For the nitric acid solution, \(1.5 = 2 \cdot 1.86 \cdot m\) gives \(m \approx 0.403 \ \mathrm{mol/kg}\). This implies both solutions have different molality due to differing \(i\) values.

Compare Other Colligative Properties

Since both solutions have equal freezing point depressions, they will also have the same boiling point elevation, vapor pressure lowering, and osmotic pressure, provided concentration and conditions remain constant, and ignoring the different nature of the solutes (non-electrolyte vs. electrolyte). However, the concentrations required to achieve this may vary due to differences in dissociation.

Key concepts

Colligative Properties

Colligative properties are unique characteristics of solutions that depend solely on the number of dissolved particles, rather than their identity. This means that it doesn't matter what the solute actually is; the key factor is how many particles it produces when dissolved. Some of the most common colligative properties include:

• Freezing point depression
• Boiling point elevation
• Vapor pressure lowering
• Osmotic pressure

These properties arise because adding solute particles to a solvent disrupts the solvent's natural tendency to form a pure solid during freezing or evaporate more readily. This shift in behavior is crucial for understanding many real-life applications, such as how antifreeze works in car engines or how salt sprinkled on icy roads helps to melt ice. In the case of both sucrose and nitric acid solutions, even though they are different substances, their common colligative property is their freezing point depression.

Molality Calculation

Molality is a measure of concentration, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on volume, molality is unaffected by temperature fluctuations, making it particularly useful for studying colligative properties like freezing point depression. To calculate molality, you can use the formula: \[ m = \frac{\Delta T_f}{i \cdot K_f} \]where

• \( \Delta T_f \) is the change in freezing point
• \( i \) is the Van't Hoff factor
• \( K_f \) is the cryoscopic constant specific to the solvent (for water, \( K_f = 1.86 \, \mathrm{°C \, kg/mol} \))

In the problem, we use this formula to determine that the molality of the sucrose solution is approximately 0.806 mol/kg while the nitric acid solution is approximately 0.403 mol/kg, showcasing how the molality differs due to their Van't Hoff factors.

Van't Hoff Factor

The Van't Hoff factor, denoted as \( i \), is a dimensionless number that describes how many particles a solute yields when it dissolves. It accounts for the effect of solute dissociation or association in solution and is critical in calculating colligative properties. The factor can differ based on whether the solute is an electrolyte or a non-electrolyte:

• For non-electrolytes, like sucrose, \( i = 1 \) because they do not dissociate into ions.
• For strong electrolytes, like nitric acid, \( i \approx 2 \) because they dissociate completely into ions (\( \mathrm{H}^+\) and \( \mathrm{NO}_3^- \)).

Understanding the Van't Hoff factor is crucial for predicting how the colligative properties will change when different substances are dissolved. In our exercise, the difference in \( i \) values between sucrose and nitric acid explains why they have different molalities when they both lower the freezing point to the same extent.

Aqueous Solutions

Aqueous solutions are those where water acts as the solvent. These solutions are incredibly common in both nature and industries due to water's excellent solvent properties. Water's polarity allows it to dissolve a wide range of substances, making aqueous solutions central to many chemical reactions, biological processes, and industrial applications. In the exercise, we're dealing with aqueous solutions of sucrose and nitric acid. While sucrose is a non-electrolyte and does not dissociate, nitric acid, as an electrolyte, splits into ions in water. This behavior is key to understanding the observed colligative properties. When considering real-life scenarios, such as cooking or chemical manufacturing, knowing whether a substance forms an aqueous solution, and the nature of that solution, is fundamental. It influences everything from flavor in cooking to reactivity in chemical processes.