Problem 46 - 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 510.3 g of NaCl to 3.0 kg of water.

Step 1: Understand the Freezing Point Depression Formula

The freezing point depression formula 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 of water (\(1.86\, \text{°C kg/mol}\)), and \( m \) is the molality of the solution.

Step 2: Calculate the Required Change in Freezing Point

The target freezing point of the solution is \(-10\, ^{\circ}C\) and the normal freezing point of water is \(0\, ^{\circ}C\). Therefore, the change in freezing point \( \Delta T_f \) is: \[ \Delta T_f = 0 - (-10) = 10\, ^{\circ}C\]

Step 3: Rearrange the Formula to Solve for Molality

Rearrange the freezing point formula to solve for the molality \( m \): \[ m = \frac{\Delta T_f}{i \cdot K_f} \] Substitute the values \( \Delta T_f = 10\, ^{\circ}C\), \( i = 1.85 \), and \( K_f = 1.86\, \text{°C kg/mol} \).

Step 4: Calculate the Molality

Substituting the values into the rearranged formula, we find the molality: \[ m = \frac{10}{1.85 \times 1.86} \approx 2.91\, \text{mol/kg} \]

Step 5: Calculate the Moles of NaCl Required

Since molality is defined as moles of solute per kilogram of solvent, multiply the molality by the mass of the water: \[ \text{moles of NaCl} = 2.91\, \text{mol/kg} \times 3.0\, \text{kg} = 8.73\, \text{mol} \]

Step 6: Convert Moles of NaCl to Mass

Use the molar mass of NaCl (approximately 58.44 g/mol) to find the mass of NaCl needed: \[ \text{mass of NaCl} = 8.73\, \text{mol} \times 58.44\, \text{g/mol} = 510.3\, \text{g} \]

Step 7: Provide the Final Answer

Therefore, to achieve a water-salt solution that freezes at \(-10\, ^{\circ}C\), you need to add approximately 510.3 grams of NaCl to 3.0 kg of water.

Key Concepts

Colligative Properties

Colligative properties are a fascinating aspect of chemistry that depend on the number of solute particles in a solution rather than the identity of the particles. This makes them particularly useful for determining how certain mixtures will behave. One of the prominent colligative properties is freezing point depression, which refers to the lowering of a solvent's freezing point upon the addition of a solute. The more solute particles present, the greater the depression of the freezing point.

In the context of making ice cream, adding rock salt to ice lowers the freezing point of the resultant ice-cream mix. This allows the milk and cream to freeze more efficiently at lower temperatures, culminating in a smoother and more delicious ice cream.

• Freezing Point Depression: The reduction in the freezing point of a liquid (solvent) caused by adding a solute.
• Boiling Point Elevation: Similar in principle, this is the increase in the boiling point.
• Osmotic Pressure and Vapor Pressure Lowering: Other examples where solute concentration affects the properties of a solution.

Van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is a critical element when it comes to understanding how solutes affect colligative properties. It represents the number of particles a solute forms in a solution. For instance, when NaCl dissolves in water, it splits into sodium ions \( (Na^+) \) and chloride ions \( (Cl^-) \), which increases the number of particles present in the solution compared to the original compound.

For NaCl, the van't Hoff factor is considered to be about 1.85, deviating slightly from the ideal value of 2 due to some ionic interactions and incomplete dissociation in real-world scenarios. This adjusted value allows for more accurate calculations when using colligative property equations.

• Dissociation in Solutions: The van't Hoff factor accounts for how solutes like NaCl actually dissociate in water.
• Effect on Freezing Point Depression: A higher van't Hoff factor generally results in a more significant freezing point depression.

Molality

Molality is a measure of the concentration of solute in a solution, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is independent of temperature because it is based on mass rather than volume, making it particularly useful for studying colligative properties that involve temperature changes.

In our ice-cream making scenario, knowing the molality helps us determine the precise concentration of NaCl needed to achieve the desired freezing point depression.

• Definition: Expressed as \( m = \frac{n}{kg} \), where \( n \) is the number of moles of solute, and \( kg \) is the mass of the solvent in kilograms.
• Consistency: Unlike molarity, molality does not change with temperature.
• Application: Used in the equation \( \Delta T_f = i \cdot K_f \cdot m \) to find the freezing point depression of a solution.