Problem 89 - Chemistry An Atoms First Approach

Question

In the winter of \(1994,\) record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of \(-29^{\circ} \mathrm{F}\) was registered. At this temperature can salting icy roads with \(\mathrm{CaCl}_{2}\) be effective in melting the ice? a. Assume \(i=3.00\) for \(\mathrm{CaCl}_{2}\) b. Assume the average value of \(i\) from Exercise 87 (The solubility of \(\mathrm{CaCl}_{2}\) in cold water is \(74.5 \mathrm{g}\) per \(100.0 \mathrm{g}\) water.

Step-by-Step Solution

Answer

In this exercise, we were asked to determine if salting icy roads with CaCl2 can be effective in melting the ice at a record low temperature of -29°C. We calculated the molality of the solution and the freezing point depression for both cases: assuming i=3.00 and assuming the average value of i from Exercise 87. For Case a, the freezing point depression is 37.5 °C, meaning that melting of the ice is effective at temperatures as low as -37.5 °C. Since the record low temperature is -29 °C, salting the icy roads with CaCl2 will be effective in melting the ice at this temperature. For Case b, without knowing the average value of i, we cannot draw a conclusion. However, the method would be the same - calculate the freezing point depression and compare it with the record low temperature.

Step 1: Calculate the molality of the solution

Since we know the solubility of CaCl2, we can calculate the molality (m) of the solution. Molality is defined as moles of solute per kilogram of solvent. First, let's find the moles of CaCl2 present in 74.5 g: Molar mass of CaCl2 = 40.08 (Ca) + 2 * 35.45 (Cl) = 110.98 g/mol Moles of CaCl2 = (74.5 g) / (110.98 g/mol) = 0.671 mol Now let's find the molality using the mass of water: Mass of water = 100 g = 0.1 kg Molality (m) = moles of solute / mass of solvent (in kg) m = 0.671 mol / 0.1 kg = 6.71 mol/kg

Step 2: Calculate freezing point depression for Case a

For Case a, we assume that i = 3.00 for CaCl2. We need the value of the molal freezing-point-depression constant (Kf) for water, which is 1.86°C/mol/kg. Now, let's calculate the freezing point depression using the formula: ∆Tf = i * Kf * m ∆Tf = 3.00 * 1.86 °C/mol/kg * 6.71 mol/kg ∆Tf = 37.5 °C

Step 3: Calculate freezing point depression for Case b

For Case b, we assume the average value of i from Exercise 87 (which is not specified here). For the sake of completeness, we will denote this average value as i_avg. The calculation will be as follows: ∆Tf = i_avg * Kf * m

Step 4: Analyze the results and draw conclusions

Now, let's analyze the results obtained from both cases. - For Case a, the freezing point depression is 37.5 °C. Since the normal freezing point of water is 0°C, this means that the melting of the ice is effective at temperatures as low as -37.5 °C. Therefore, since the record low temperature is -29 °C, salting the icy roads with CaCl2 will be effective in melting the ice at this temperature. - For Case b, we cannot conclude without knowing the average value of i. However, the method would be the same - calculate the freezing point depression and compare it with the record low temperature to determine the effectiveness of CaCl2 in melting the ice at -29 °C.

Key Concepts

Molality

Molality is a concentration unit expressing the amount of solute per kilogram of solvent.
It is especially useful for temperature-dependent experiments, like freezing point depression.
Understanding molality helps evaluate how effectively a solute can alter the freezing point.
To calculate molality, first determine the moles of the solute,

• Take the mass of the solute (in grams).
• Divide it by its molar mass (g/mol) to convert to moles.

For example, with 74.5 grams of calcium chloride (CaCl₂), and its molar mass being 110.98 g/mol,

• Moles = 74.5 g / 110.98 g/mol = 0.671 mol.

Next, convert the mass of the solvent to kilograms, here it's 0.1 kg of water. Then calculate molality as:

• Molality (m) = 0.671 mol / 0.1 kg = 6.71 mol/kg.

This measure allows you to explore how substances like calcium chloride influence the freezing point.

Colligative Properties

Colligative properties depend solely on solute quantity, not its type.
These properties include boiling point elevation and freezing point depression.
One vital concept is freezing point depression:

• The freezing point of a solvent is lowered by adding a solute.
• This process is exploited for de-icing roads using substances like calcium chloride (CaCl₂).

For calculations, use the expression: \[ \Delta T_f = i \times K_f \times m \]Where - \( i \) is the van 't Hoff factor, indicating solute dissociation level,- \( K_f \) is the freezing-point-depression constant of the solvent,- and \( m \) is the molality calculated as in the previous section.In the exercise, the van 't Hoff factor \( i \) for CaCl₂ is considered 3.0, assuming it fully dissociates into three ions per formula unit.Applying it:

• For calcium chloride and water, with a \( K_f \) of 1.86°C/mol/kg,
• ∆\( T_f = 3.0 \times 1.86 \times 6.71 \approx 37.5 °C \)

This result indicates a potential to lower water's freezing point from 0°C to -37.5°C.

Calcium Chloride Application

Calcium chloride is commonly applied in de-icing due to its strong colligative effects.
When dispersed on icy roads, it effectively lowers the freezing point. This creates a saltwater solution that remains liquid at lower temperatures.
Benefits include:

• Efficient at low temperatures, outperforming other salts like sodium chloride.
• Works even when temperatures plunge to around -29°C or lower.
• Increases road safety by reducing ice formation.

This application is based on scientific principles of colligative properties.
The effectiveness relies significantly on factors like proper concentration and accurate predictions of ambient temperatures.
Using calcium chloride in cold climates helps ensure melted ice doesn't refreeze, largely due to its substantial freezing point depression capabilities.