Problem 87 - Chemistry: An Atoms-Focused Approach

Question

Melting Ice \(\mathrm{CaCl}_{2}\) is used to melt ice on sidewalks. Could \(\mathrm{CaCl}_{2}\) melt ice at \(-20^{\circ} \mathrm{C} ?\) Assume that the solubility of \(\mathrm{CaCl}_{2}\) at this temperature is \(70.1 \mathrm{g} \mathrm{CaCl}_{2} / 100.0 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) and that the van 't Hoff factor for a saturated solution of \(\mathrm{CaCl}_{2}\) is 2.5.

Step-by-Step Solution

Answer

Answer: ________________________ (Calculate the freezing point depression using the given information and compare it to 20°C. If it is greater than 20°C, CaCl2 can indeed melt ice at -20°C.)

Step 1: Calculate the molality of the CaCl2 solution

To find the freezing point depression, we need the molality of the CaCl2 solution. Molality (m) is defined as moles of solute per kilogram of solvent. Given solubility: 70.1g CaCl2 / 100.0g H2O Molar mass of CaCl2 = 40.08(Ca) + 35.45(Cl) * 2 = 110.98 g/mol Molality (m) = (moles of CaCl2) / (mass of solvent in kg) Molality (m) = (70.1 g CaCl2 / 110.98 g/mol) / (100.0 g H2O / 1000)

Step 2: Calculate the freezing point depression

To find the freezing point depression, we use the formula: ΔTf = i * Kf * m where ΔTf is the freezing point depression, i is the van 't Hoff factor, Kf is the freezing point depression constant for water (1.86°C·kg/mol), and m is the molality of the solution. Given i = 2.5, we can calculate ΔTf: ΔTf = (2.5) * (1.86°C·kg/mol) * m

Step 3: Compare the freezing point depression to the temperature difference

Now, we need to compare the calculated freezing point depression with the temperature difference between the usual freezing point of water (0°C) and -20°C: Temperature difference = 20°C If the calculated freezing point depression is greater than 20°C, CaCl2 can melt ice at -20°C.

Step 4: Conclusion

Calculate the freezing point depression using the given information and compare it to 20°C. If it is greater than 20°C, CaCl2 can indeed melt ice at -20°C.

Key Concepts

Solubility

Solubility is a measure of how much solute can be dissolved in a certain amount of solvent at a specific temperature. It's often expressed in terms of grams of solute per 100 grams of solvent. In our example,

• The solubility of calcium chloride (\(\text{CaCl}_2\)) at \(-20^{\circ} \text{C}\) is given as 70.1 grams per 100 grams of water.

This implies that at this temperature, 70.1 grams of \(\text{CaCl}_2\) can be dissolved in every 100 grams of water to form a saturated solution.

Understanding solubility is crucial because it helps us know how much \(\text{CaCl}_2\) is available to interact with the ice. This interaction, as we'll see, is important for lowering the freezing point of water and thereby melting the ice.

Different substances have different solubility levels depending on temperature, and for effective ice melting the solute should readily dissolve at low temperatures.

Molality

Molality is a concentration term that describes the number of moles of solute per kilogram of solvent. Unlike molarity, which is affected by changes in temperature and volume, molality remains constant because it is based on mass rather than volume.

To calculate the molality (\(m\)),

• we use the formula: \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \).

For \(\text{CaCl}_2\), given that we have 70.1 grams of its solubility per 100 grams of water, we first need its molar mass, which is calculated as:

• Molar mass of \(\text{CaCl}_2 = 110.98 \text{ g/mol}\).

Then, convert the mass of the water into kilograms (100 g = 0.1 kg) and find the moles of \(\text{CaCl}_2\) in the 70.1 g. Through these conversions, we'll get the molality, a key player in determining how much the freezing point is depressed.

van 't Hoff factor

The van 't Hoff factor (\(i\)) is an important dimensionless number used in colligative properties. It represents the number of particles the solute splits into when dissolved.

In our problem, the van 't Hoff factor for the solution of \(\text{CaCl}_2\) is mentioned as 2.5. Theoretically, \(\text{CaCl}_2\) should dissociate into three ions: one \(\text{Ca}^{2+}\) ion and two \(\text{Cl}^-\) ions. This provides an ideal factor of 3. However, the experimental factor of 2.5 suggests incomplete dissociation.

The van 't Hoff factor plays a crucial role in calculating the freezing point depression (\(\Delta T_f\)), as seen in the equation: \(\Delta T_f = i \cdot K_f \cdot m\). Given \(i = 2.5\), it implies that the \(\text{CaCl}_2\) solution, when used, affects the freezing point of water by a factor of 2.5, which is crucial in predicting whether \(\text{CaCl}_2\) can melt ice effectively.