Question

The solubility of NaCl in water at \(100^{\circ} \mathrm{C}\) is \(39.1 \mathrm{g} / 100 .\) g of water. Calculate the boiling point of this solution. (Assume \(i=1.85\) for \(\mathrm{NaCl}.\)

Short answer

The boiling point of the NaCl solution is approximately 106.33 °C.

Step-by-step solution

Understand the Problem

We are tasked to calculate the boiling point of a solution of NaCl in water at a high concentration, given the solubility of NaCl and its van't Hoff factor \(i\). We will use the boiling point elevation formula to determine the new boiling point.

Use Boiling Point Elevation Formula

The formula for boiling point elevation is \(\Delta T_b = i \cdot K_b \cdot m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor (1.85), \(K_b\) is the ebullioscopic constant of water \(0.512 \, \text{°C/m}\), and \(m\) is the molality of the solution.

Calculate Molality of the Solution

The solubility is given as 39.1 g of NaCl per 100 g of water. Convert this mass to moles by dividing by the molar mass of NaCl \((58.44\ g/mol)\). Then divide by the mass of water in kilograms to find molality. \[ m = \frac{39.1\, \text{g} / 58.44\, \text{g/mol}}{0.1\, \text{kg}} \approx 6.69 \text{ mol/kg} \]

Calculate Boiling Point Elevation

Plug the calculated molality into the formula for boiling point elevation: \[ \Delta T_b = 1.85 \cdot 0.512 \, \text{°C/m} \cdot 6.69 \, \text{mol/kg} = 6.33 \, \text{°C} \]

Determine New Boiling Point

Add the boiling point elevation \(\Delta T_b\) to the normal boiling point of water \(100 \text{°C}\) to find the new boiling point. \[ 100 \, \text{°C} + 6.33 \, \text{°C} = 106.33 \, \text{°C} \]

Key concepts

Solubility of NaCl

Solubility refers to the ability of a substance to dissolve in a solvent, such as NaCl (table salt) dissolving in water. At a specific temperature, solubility dictates how much of a substance can be dissolved in a given amount of solvent before reaching saturation. In this case, the solubility of NaCl at 100°C is given as 39.1 grams per 100 grams of water.

This means if you have 100 grams of water heated to 100°C, you can dissolve up to 39.1 grams of NaCl before the solution becomes saturated. Any additional salt will not dissolve unless more water is added or temperature changes. Understanding solubility is crucial for calculating the concentration of solutions and determining physical properties like boiling and freezing points.

van't Hoff factor

The van't Hoff factor (\(i\)) is an important concept in colligative properties, which are properties that depend on the number of particles in a solution, rather than their identity. It represents the number of particles into which a compound dissociates in solution.

For NaCl, which dissociates into sodium (\(\text{Na}^+\)) and chloride (\(\text{Cl}^-\)) ions, you'd expect the van't Hoff factor to be around 2, since each molecule splits into two particles. However, in practice, the value can be slightly lower due to ion pairing or incomplete dissociation. In this exercise, the van't Hoff factor is given as 1.85, indicating some ion interaction occurs, affecting how the ions behave in the solution.

Molality

Molality (\(m\)) is a concentration term used primarily in colligative properties calculations. It is defined as the number of moles of solute dissolved in one kilogram of solvent.

To find the molality of our NaCl solution, we must first convert the mass of solute (NaCl) to moles, using its molar mass. Then, divide this by the mass of the solvent expressed in kilograms. Molality is advantageous since it does not change with temperature, unlike molarity, which involves volume.

• Given: 39.1 g NaCl per 100 g water
• Molar Mass of NaCl: 58.44 g/mol
• Convert mass of water to kg: 100 g = 0.1 kg
• Calculate: \(\text{molality} = \frac{39.1 \, \text{g} / 58.44 \, \text{g/mol}}{0.1 \, \text{kg}} \approx 6.69 \, \text{mol/kg}\)

Ebullioscopic Constant

The ebullioscopic constant (\(K_b\)) is a property of the solvent, used to calculate boiling point elevation in solutions. It is expressed in °C per molal (\(° ext{C/m}\)). Different solvents have different ebullioscopic constants, reflecting how they affect the boiling point when a solute is added.

For water, the ebullioscopic constant is 0.512 °C/m. This constant helps quantify the degree of boiling point elevation when a non-volatile solute like NaCl is dissolved in water. The increase in boiling point \(\Delta T_b\) is determined through the formula \(\Delta T_b = i \cdot K_b \cdot m\), where all these values are plugged in to find by how many degrees the boiling point is increased.