Question

The organic salt \(\left[\left(\mathrm{C}_{4} \mathrm{H}_{9}\right)_{4} \mathrm{~N}\right]\left[\mathrm{ClO}_{4}\right]\) consists of the ions \(\left(\mathrm{C}_{4} \mathrm{H}_{9}\right)_{4} \mathrm{~N}^{+}\) and \(\mathrm{ClO}_{4}^{-}\). The salt dissolves in chloroform. What mass (in grams) of the salt must have been dissolved if the boiling point of a solution of the salt in \(25.0 \mathrm{~g}\) chloroform is \(63.20^{\circ} \mathrm{C} ?\) The normal boiling point of chloroform is \(61.70{ }^{\circ} \mathrm{C}\) and \(K_{\mathrm{b}}=3.63{ }^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\). Assume that the salt dissociates completely into its ions in solution.

Short answer

1.25 grams of the salt must have been dissolved.

Step-by-step solution

Calculate Boiling Point Elevation

The boiling point elevation (\( \Delta T_b \)) is the difference between the boiling point of the solution and the normal boiling point of chloroform. This can be calculated as: \[\Delta T_b = 63.20^{\circ} \mathrm{C} - 61.70{ }^{\circ} \mathrm{C} = 1.50^{\circ} \mathrm{C}\]

Apply Boiling Point Elevation Formula

The boiling point elevation is given by \( \Delta T_b = i \cdot K_b \cdot m \), where \( i \) is the van 't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality. Since the salt dissociates into 2 ions, \( i = 2 \).Now solve for \( m \): \[1.50 = 2 \cdot 3.63 \cdot m\]\[m = \frac{1.50}{2 \cdot 3.63}\]\[m \approx 0.2067 \ \mathrm{mol/kg}\]

Calculate Moles of Salt

Molality \( m \) is defined as moles of solute per kg of solvent. Thus, the moles of salt can be calculated as:\[\text{Moles of salt} = m \cdot \text{kg of chloroform}\]Convert 25.0 g chloroform to kg:\[25.0 \text{ g} = 0.025 \text{ kg}\]Now calculate moles of salt:\[\text{Moles of salt} = 0.2067 \cdot 0.025 \approx 0.0051675 \ \mathrm{mol}\]

Calculate Mass of Salt

Calculate the molar mass of the salt \( \left[\left(\mathrm{C}_4 \mathrm{H}_9\right)_4 \mathrm{N}\right]\left[\mathrm{ClO}_4\right] \):- \( \left(\mathrm{C}_4 \mathrm{H}_9\right)_4 \mathrm{N}^+\): - Carbon (C): 12.01 g/mol \(\times 16\) - Hydrogen (H): 1.01 g/mol \(\times 36\) - Nitrogen (N): 14.01 g/mol \(\times 1\)- \( \mathrm{ClO}_4^- \): - Chlorine (Cl): 35.45 g/mol - Oxygen (O): 16.00 g/mol \(\times 4\)Molar mass of salt = 242.43 g/mol. Now calculate the mass:\[\text{Mass of salt} = \text{Moles of salt} \times \text{Molar mass of salt}\]\[\text{Mass of salt} = 0.0051675 \times 242.43 \]\[\text{Mass of salt} \approx 1.25 \ \mathrm{g}\]

Key concepts

Molality

Molality is a measure of concentration used in chemistry to express the amount of solute present in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is independent of temperature and pressure because it is based on the mass of the solvent, not its volume.
In the boiling point elevation problem we are focused on, molality helps us determine how the concentration of the dissolved organic salt affects the boiling point of a solvent—in this case, chloroform.
The formula to calculate molality (\(m\)) is:\[m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\]Calculating molality is a crucial step in finding out how much the boiling point of the solvent has increased due to the presence of a solute. This is used in the formula for boiling point elevation, \(\Delta T_b = i \cdot K_b \cdot m\), which combines molality with other factors to find the change in boiling point.

Van 't Hoff factor

The Van 't Hoff factor,\(i\), is an important concept in understanding how many particles a solute dissociates into when it dissolves in a solvent. For non-dissociating solutes, \(i\) is usually 1, while for those that dissociate completely, \(i\) represents the number of particles into which a single solute unit divides.
In our given problem, the organic salt dissociates into two distinct ions—\(\left(\mathrm{C}_4 \mathrm{H}_9\right)_4 \mathrm{N}^+\) and \(\mathrm{ClO}_4^-\)—therefore, \(i = 2\). This factor is used in the boiling point elevation formula to account for the increased number of solute particles in the solution compared to what is suggested by the concentration of the compound before dissociation.

• The Van 't Hoff factor is crucial for calculating how properties like boiling point and freezing point are affected by the dissolved solute.
• It represents the ratio between the actual number of particles after dissolution and the number of formula units initially dissolved.

Understanding the Van 't Hoff factor allows us to accurately predict physical changes in solutions when solutes are added.

Ebullioscopic constant

The ebullioscopic constant,\(K_b\), is a property specific to each solvent that quantifies how much the boiling point rises per molal concentration of a solute. Each solvent has a unique \(K_b\) value, making it essential for the calculation of boiling point elevation.
The ebullioscopic constant is part of the formula used to determine the boiling point elevation,\(\Delta T_b = i \cdot K_b \cdot m\). In this formula:

• \(\Delta T_b\) is the change in boiling point.
• \(i\) is the Van 't Hoff factor.
• \(m\) is the molality of the solution.

The ebullioscopic constant is used to calculate how much the solvent's boiling point will increase when a solute is added. For the calculation given in the exercise, chloroform has an ebullioscopic constant of \(3.63 \ ^\circ \mathrm{C} \cdot \mathrm{kg} \cdot \mathrm{mol}^{-1}\). This value helps us relate the molality of the solution and the dissociation of the solute (the Van 't Hoff factor) to the overall boiling point elevation.