Problem 48 - Chemistry and Chemical Reactivity

Question

To make homemade ice cream, you cool the milk and cream by immersing the container in ice and a concentrated solution of rock salt (NaCl) in water. If you want to have a water-salt solution that freezes at \(-10 .^{\circ} \mathrm{C},\) what mass of NaCl must you add to \(3.0 \mathrm{kg}\) of water? (Assume the van't Hoff factor, \(i,\) for \(\mathrm{NaCl}\) is \(1.85 .\) )

Step-by-Step Solution

Answer

Add approximately 510 g of NaCl to 3 kg of water.

Step 1: Understand Freezing Point Depression

Freezing point depression is a colligative property, where the addition of a solute lowers the freezing point of a solvent. The equation to calculate freezing point depression is given by: \[ \Delta T_f = i \cdot K_f \cdot m \]where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant (for water, \( K_f = 1.86 \, \text{°C kg/mol} \)), and \( m \) is the molality of the solution.

Step 2: Calculate Freezing Point Depression Required

We need the solution to freeze at \(-10 \, ^{\circ}\text{C}\). Water normally freezes at \(0 \, ^{\circ}\text{C}\), so the required change in freezing point is:\[ \Delta T_f = 0 - (-10) = 10 \, ^{\circ}\text{C} \]

Step 3: Rearrange the Freezing Point Depression Formula

Using the freezing point depression equation \( \Delta T_f = i \cdot K_f \cdot m \), we rearrange to solve for molality:\[ m = \frac{\Delta T_f}{i \cdot K_f} \]

Step 4: Calculate Molality of Solution

Substitute the known values into the rearranged equation:\[ m = \frac{10 \, ^{\circ}\text{C}}{1.85 \times 1.86 \, ^{\circ}\text{C kg/mol}} \]Find \( m \):\[ m = \frac{10}{3.441} \approx 2.91 \, \text{mol/kg} \]

Step 5: Calculate Moles from Molality

Molality \( (m) \) is the moles of solute per kilogram of solvent. With \(3.0 \, \text{kg}\) of water, calculate the moles of NaCl:\[ \text{moles of NaCl} = m \times \text{kg of solvent} = 2.91 \, \text{mol/kg} \times 3.0 \, \text{kg} = 8.73 \, \text{mol} \]

Step 6: Convert Moles of NaCl to Mass

The molar mass of NaCl is approximately \(58.44 \, \text{g/mol}\). Convert moles of NaCl to grams:\[ \text{mass of NaCl} = 8.73 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 510 \, \text{g} \]

Key Concepts

Colligative Properties

Colligative properties are fascinating aspects of solutions because they depend purely on the number of solute particles and not the specific type of particles. These include properties like freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure. For a given solvent, adding any solute can affect its physical properties.

In the scenario of making homemade ice cream, we take advantage of one such colligative property - freezing point depression. When you add a solute like salt to ice, it lowers the temperature at which ice can freeze. The extent of this drop in freezing point is determined by the quantity of solute particles in the solution.

• The more particles, the greater the depression in the freezing point.
• Colligative properties are independent of the solute's identity, only the particle count matters.

Understanding colligative properties helps us manipulate conditions like freezing and boiling for various practical applications.

Van't Hoff Factor

The van't Hoff factor, denoted as \(i\), is a measure of the effect a solute has on colligative properties. It represents the number of particles each mole of solute contributes when dissolved in a solution. For example, NaCl dissociates into Na⁺ and Cl⁻ ions, suggesting it could double the number of effective particles compared to non-dissociating solutes.

However, interactions in solutions can slightly alter expected dissociation. The van't Hoff factor helps adjust for this, with the provided example having \(i = 1.85\) for NaCl. This indicates that in reality, each formula unit of NaCl produces about 1.85 particles, slightly less than its theoretical maximum of 2. Understanding \(i\) helps accurately predict changes in properties like freezing and boiling points.

• An \(i\) of 1 suggests no dissociation or association.
• Values greater than 1 indicate dissociation into ions.

Cryoscopic Constant

The cryoscopic constant, \(K_f\), is a property that characterizes how much the freezing point of a solvent will decrease for each mole of solute particles per kilogram of solvent. It's a crucial component in calculating freezing point depression. Each solvent has its own unique \(K_f\).

For water, which is commonly used, \(K_f = 1.86 \, \text{°C kg/mol}\). This constant allows us to determine how strong an effect a given amount of solute will have on lowering the freezing point of water. Knowing this constant is essential because it bridges the relationship between particle concentration and the degree of freezing point depression.

• A higher \(K_f\) means a more significant freezing point depression per mole of solute.
• For accuracy, \(K_f\) needs precise measurement and often courtesy of empirical data.

Molality

Molality is a concentration unit defined as moles of solute per kilogram of solvent. Unlike molarity, which depends on volume, molality is temperature-independent because it uses mass. This makes it especially useful in studying colligative properties, as they rely strictly on the concentration of particles regardless of temperature changes.

In the ice cream making process, knowing that the solution's molality helps determine the amount of solute needed to achieve a specific freezing point change. The calculation involves dividing the moles of solute by the mass (in kg) of the solvent.

• Molality ensures precise measurement unaffected by thermal expansion or contraction.
• For colligative properties, molality often provides more consistent and accurate results than molarity.

This precision is why molality is the preferred metric for calculating such properties as freezing point depression.