Question

An aqueous solution containing \(0.250\) mole of \(\mathrm{Q}\), a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{~g}\) water freezes at \(-2.79^{\circ} \mathrm{C}\). What is the van't Hoff factor for Q? The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\). What is the formula of \(\mathrm{Q}\) if it is \(38.68 \%\) chlorine by mass and there are twice as many anions as cations in one formula unit of \(\mathrm{Q}\) ?

Short answer

The van't Hoff factor for compound Q is found to be 3. Given that Q is 38.68% chlorine by mass and has twice as many anions as cations, the molar mass of the cation is calculated to be 167.95 g/mol which is closest to the molar mass of Pb²⁺. Therefore, the formula of compound Q is \(Pb_2Cl_{12}\) or \(Pb_2(Cl_6)\).

Step-by-step solution

Calculate the molality of the solution

First, let's find the molality (moles of solute per kilogram of solvent) of the solution. We are given 0.250 moles of Q dissolved in 5.00 × 10² g of water. Molality (m) = (moles of solute) / (mass of solvent in kg) The mass of water is given in grams, so we need to convert it to kilograms: Mass of water = (5.00 × 10² g) × (1 kg / 1000 g) = 0.500 kg Now, we can calculate the molality: m = (0.250 moles) / (0.500 kg) = 0.500 mol/kg

Calculate the van't Hoff factor using freezing-point depression

The freezing-point depression ΔTf can be expressed by the equation: ΔTf = i × Kf × m where i is the van't Hoff factor, Kf is the molal freezing-point depression constant for water, and m is the molality of the solution. We are given ΔTf = -2.79°C (as a negative value because it is a depression) and Kf = 1.86°C⋅kg/mol. Using these values and the molality we found in Step 1, we can solve for the van't Hoff factor i: -2.79°C = i × (1.86°C⋅kg/mol) × (0.500 mol/kg) First, let's divide both sides of the equation by (1.86°C⋅kg/mol) × (0.500 mol/kg): i = -2.79°C / [(1.86°C⋅kg/mol) × (0.500 mol/kg)] = -3.0 Thus, the van't Hoff factor i = 3.

Determine the formula of compound Q

We are given that compound Q is 38.68% chlorine by mass. Let the molar mass of the cation in Q be M_cation and the molar mass of Cl be M_Cl (35.45 g/mol). We know there are twice as many anions as cations in one formula unit of Q, so the formula can be written as Q_m = (Cation)₂╳(Anion₆). The mass ratio of chlorine in the compound is: (6 × M_Cl) / [(2 × M_cation) + (6 × M_Cl)] = 0.3868 By simplifying the above equation, we can find the ratio of M_cation to M_Cl: (2 × M_cation) / M_Cl = 9.468 Now substituting M_Cl = 35.45 g/mol in the equation: 2 × M_cation = 9.468 × 35.45 = 335.9 g/mol So the molar mass of the cation (M_cation) = 335.9 / 2 = 167.95 g/mol The molar mass of the cation is closest to the molar mass of Pb²⁺, which is 207.2 g/mol (the actual value can be slightly different from the periodic table values). So, the cation in Q is Pb²⁺ and thus, the formula of compound Q is: Q = Pb₂(Cl₆) or Q = Pb₂Cl₁₂

Key concepts

Molal Freezing-Point Depression

Molal freezing-point depression is a phenomenon that occurs when a solute is added to a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. This is due to the presence of solute particles interfering with the formation of a solid lattice structure, thereby requiring a lower temperature to solidify.

The molal freezing-point depression can be calculated using the formula: \[ \Delta T_f = i \times K_f \times m \]where:

• \( \Delta T_f \) is the freezing-point depression (the change in freezing point),
• \( i \) is the van't Hoff factor (which indicates the number of particles into which a compound dissociates in solution),
• \( K_f \) is the molal freezing-point depression constant specific to the solvent (for water, it's \(1.86\, ^\circ \text{C} \cdot \text{kg/mol}\)), and
• \( m \) is the molality of the solution in \( \text{mol/kg} \).

As seen in the problem, the calculated molality was 0.500 mol/kg, and the resulting freezing-point change was -2.79°C. Using these values helps us to determine the van't Hoff factor, which reflects the degree of dissociation of the electrolyte in solution.

Strong Electrolyte

A strong electrolyte is a compound that completely ionizes in solution. This means that when dissolved, it dissociates into its constituent ions, resulting in a solution that can conduct electricity efficiently.

In the context of the problem, compound \( \text{Q} \) is identified as a strong electrolyte because it dissociates into multiple ions, as indicated by its van't Hoff factor of 3. This factor tells us that the substance produces three ions per formula unit in solution.

Examples of strong electrolytes include:

• Strong acids like \( \text{HCl} \)
• Strong bases like \( \text{NaOH} \)
• Salts such as \( \text{NaCl} \)

Understanding strong electrolytes is crucial in predicting the colligative properties of solutions, such as freezing point depression, boiling point elevation, and osmotic pressure. Each of these properties depends on the number of particles in the solution, which can significantly affect the behavior of the solution.

Chlorine Mass Percentage

Chlorine mass percentage in a compound is a measure of how much of the compound's mass is made up of chlorine atoms. It is expressed as a percentage, showing the proportion of chlorine relative to the total mass of the molecule.

To find the chlorine mass percentage, use the formula:\[ \text{Chlorine Mass Percentage} = \frac{\text{Mass of chlorine}}{\text{Total mass of compound}} \times 100\% \]
For compound \( \text{Q} \), which has a chlorine mass percentage of 38.68%, the configuration involves twice as many anions (which are chlorine ions) as cations. This means that in one formula unit of \( \text{Q} \), the chlorine contributes significantly to its mass. By determining the mass ratio, we can then solve for the molar mass of the cation involved, aiding in deducing the formula of the compound.

This concept is crucial when dealing with complex reactions or mixtures, as it allows chemists to determine the exact contribution of different elements within compounds.