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 \(\mathrm{Q}\) ?

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

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.

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