The compound \(\mathrm{VCl}_{4}\) undergoes dimerization in solution: $$ 2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8} $$ When \(6.6834 \mathrm{~g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{~g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(1.696\) \(\mathrm{g} / \mathrm{cm}^{3}\), and \(K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\) for \(\mathrm{CCl}_{4} .\).

The first step is to calculate the molality (m) of the VCl4 solution using the freezing point depression formula: ΔTf = Kf * m where ΔTf is the freezing point depression, Kf is the cryoscopic constant of CCl4, and m is the molality of the VCl4 solution. We can rearrange the formula to solve for molality: m = ΔTf / Kf = 5.97 °C / 29.8 °C kg/mol = 0.200 mol/kg

Now we can determine the moles of VCl4 dissolved in the solution: mass of VCl4 = 6.6834 g molecular weight of VCl4 = (50.94 + 4*35.45) g/mol = 192.74 g/mol moles of VCl4 = (6.6834 g) / (192.74 g/mol) = 0.0347 mol Using the molality value from step 1, we can calculate the molality of VCl4 in terms of moles of VCl4 and the mass of the solvent (CCl4): molality = (moles of VCl4) / (mass of CCl4 in kg) => 0.200 mol/kg = (0.0347 mol) / (mass of CCl4 in kg) mass of CCl4 in kg = 0.0347 mol / 0.200 mol/kg = 0.1735 kg = amount of CCl4 in the solution As the equilibrium concentration will be affected by initial concentration and conversion, let x be the molality of V2Cl8 formed at equilibrium. Therefore, the molality of VCl4 at equilibrium will be (0.200 - x) mol/kg.

At equilibrium, we can apply the freezing point depression formula again taking into account the contributions of both VCl4 and V2Cl8: ΔTf = Kf * (molality of VCl4 + 2 * molality of V2Cl8) 5.97 = 29.8 (0.200 - x + 2x) Now, we can solve for x: 5.97 / 29.8 = 0.200 + x x = 0.001 mol/kg

Finally, we can find the equilibrium constant (Kc) for the dimerization reaction. The reaction is given by: 2 VCl4 <=> V2Cl8 We can write the equilibrium expression for this reaction: Kc = [V2Cl8] / [VCl4]^2 First, we need to find the molar concentration of VCl4 and V2Cl8 at equilibrium. We will convert their respective molalities to molar concentrations using the density of the equilibrium mixture: Equilibrium mixture density = 1.696 g/cm³ Total volume of the mixture = [(mass of VCl4 + mass of CCl4)] / (density of equilibrium mixture) Total volume of the mixture = (6.6834 g + 100.0 g) / (1.696 g/cm³) = 62.917 cm³ Convert the volume to liters: Total volume of the mixture = 62.917 cm³ * (1 L / 1000 cm³) = 0.062917 L Now, we can find the molar concentrations: [V2Cl8] = (0.001 mol/kg) * (0.1735 kg) / (0.062917 L) = 0.00275 mol/L [VCl4] = (0.200 mol/kg - 0.001 mol/kg) * (0.1735 kg) / (0.062917 L) = 0.551 mol/L Now, we can substitute these equilibrium concentrations into the Kc expression: Kc = (0.00275 mol/L) / (0.551 mol/L)^2 Kc ≈ 0.0090 The equilibrium constant for the dimerization of VCl4 at the given temperature is approximately 0.0090.