Problem 63 - Chemistry and Chemical Reactivity

Question

Instead of using NaCl to melt the ice on your sidewalk, you decide to use \(\mathrm{CaCl}_{2}\). If you add \(35.0 \mathrm{g}\) of \(\mathrm{CaCl}_{2}\) to 150. g of water, what is the freezing point of the solution? (Assume \(i=2.7\) for \(\mathrm{CaCl}_{2 .}\) )

Step-by-Step Solution

Answer

The freezing point of the solution is approximately \(-10.55~\mathrm{°C}\).

Step 1: Identify the Given Information

First, let's identify all the given information from the problem:- Mass of \(\mathrm{CaCl}_2\) is \(35.0~\mathrm{g}\).- Mass of water is \(150.0~\mathrm{g}\).- Van't Hoff factor \((i)\) for \(\mathrm{CaCl}_2\) is \(2.7\).We need to find the freezing point of the solution.

Step 2: Calculate the Number of Moles of CaCl2

Calculate the moles of \(\mathrm{CaCl}_2\) using its molar mass.The molar mass of \(\mathrm{CaCl}_2\) is approximately \(110.98 ~ \mathrm{g/mol}\).\[\text{Moles of } \mathrm{CaCl}_2 = \frac{35.0 ~ \mathrm{g}}{110.98 ~ \mathrm{g/mol}} \approx 0.315~\mathrm{moles}\]

Step 3: Calculate Molality\((m)\) of the Solution

Molality \((m)\) is calculated using the formula:\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\]Mass of water in kg = \(150.0 ~\mathrm{g} = 0.150 ~\mathrm{kg}\)\[m = \frac{0.315~\mathrm{moles}}{0.150~\mathrm{kg}} \approx 2.10 ~\mathrm{mol/kg}\]

Step 4: Calculate the Freezing Point Depression \((\Delta T_f)\)

The freezing point depression \((\Delta T_f)\) is calculated using the formula:\[\Delta T_f = i \cdot K_f \cdot m\]Where the cryoscopic constant \(K_f\) for water is approximately \(1.86 ~\mathrm{°C/m}\).\[\Delta T_f = 2.7 \times 1.86 ~\mathrm{°C/m} \times 2.10 ~\mathrm{mol/kg} \approx 10.55 ~\mathrm{°C}\]

Step 5: Calculate the New Freezing Point

The new freezing point of water is the normal freezing point minus the freezing point depression:Normal freezing point of water = \(0.0 ~\mathrm{°C}\).\[\text{Freezing point of solution} = 0.0 ~\mathrm{°C} - 10.55 ~\mathrm{°C} = -10.55 ~\mathrm{°C}\]

Key Concepts

Colligative Properties

Colligative properties are fascinating because they depend only on the number of solute particles in a solution, not the type of particles. This means properties like boiling point elevation and freezing point depression don't care whether you're dissolving sugar or salt; they just care about how many particles are in the solution.

In the case of freezing point depression, the addition of solute particles disrupts the crystal formation, meaning the solution must be cooled to a lower temperature than pure solvent before freezing occurs. This is why adding a substance like \(\text{CaCl}_2\) to water lowers the freezing point, helping melt ice on sidewalks in winter.

Understanding colligative properties is essential for many real-world applications, such as antifreeze in car engines. It is also crucial in calculating the freezing point depression for the exercise at hand.

Van't Hoff Factor

The Van't Hoff factor \((i)\) is an important concept for understanding how solutions behave. It accounts for the actual number of particles into which a solute dissociates in solution. For ionic compounds like \(\text{CaCl}_2\), the Van't Hoff factor is particularly relevant because it dissociates into more than one ion.

For \(\text{CaCl}_2\), it dissociates into three ions: one \(\text{Ca}^{2+}\) ion and two \(\text{Cl}^-\) ions. However, due to interactions in solution, the effective Van't Hoff factor isn't always an integer. In this example, \(i\) is given as 2.7, reflecting these real-world interactions.

This factor is crucial for calculating the freezing point depression since it multiplies with the molality and the cryoscopic constant to find the final value. Understanding \((i)\) helps in predicting and calculating the effects a solute will have on the freezing point.

Molality

Molality \((m)\) is a measure of the concentration of a solute in a solvent, expressed in moles of solute per kilogram of solvent. It's often used in colligative properties calculations because it doesn’t change with temperature, unlike molarity.

For our example, we first calculated the moles of \(\text{CaCl}_2\) using its mass divided by its molar mass. Then, with water as the solvent, we converted its mass from grams to kilograms for use in our formula.

The formula for molality is simple:

• \(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)

In this case, the calculated molality is about 2.10 mol/kg. Having this value is necessary for determining how much the freezing point will be lowered by the addition of \(\text{CaCl}_2\). Understanding molality is vital in scenarios where temperature fluctuations are involved.