In a solution with carbon tetrachloride as the solvent, the compound \(\mathrm{VCl}_{4}\) undergoes dimerization: $$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When 6.6834 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 \(\mathrm{CCl}_{4} . )\)

We know that the freezing point depression is given by the formula: $$ΔT_f = K_f × \text{molality}$$ Where ΔT_f is the freezing point depression, K_f is the cryoscopic constant for the solvent, and the molality is the concentration of the solute. We are given ΔT_f = 5.97°C and K_f = 29.8°C kg/mol, so we can solve for the molality: $$\text{molality} = \dfrac{ΔT_f}{K_f} = \dfrac{5.97\,^{\circ}\text{C}}{29.8\,^{\circ}\text{C·kg/mol}} = 0.2 \text{mol/kg (carbon tetrachloride)}$$

First, find the moles of VCl4 in the solution: We are given 6.6834 g of VCl4, and the molar mass of VCl4 = 152.88 g/mol $$\text{moles}_{VCl4} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{6.6834\,\text{g}}{152.88\,\text{g/mol}} = 0.0437\, \text{mol}$$ Next, find the moles of carbon tetrachloride solvent: Given 100 g of solvent and the molar mass of carbon tetrachloride = 153.82 g/mol $$\text{moles}_{CCl4} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{100\,\text{g}}{153.82\,\text{g/mol}} = 0.65\, \text{mol}$$

Now, let us find the initial concentration of the two compounds in the equilibrium equation using the volume of the solution. The volume can be calculated using the density of the solution. $$\text{mass}_{total} = \text{mass}_{VCl4} + \text{mass}_{CCl4} = 6.6834\,\text{g} + 100\,\text{g} = 106.6834\,\text{g}$$ Given, density \(=1.696\, \text{g/cm}^{3}\) \(\implies\) Volume = \(\dfrac{\text{mass}_{total}}{\text{density}} = \dfrac{106.6834\,\text{g}}{1.696\, \text{g/cm}^{3}} = 62.92\,\text{cm}^{3} = 62.92 \times 10^{-3}\, \text{L}\) Now, we can find the initial concentration of VCl4 and V2Cl8: $$[\text{VCl4}]_{initial} = \dfrac{\text{moles}_{VCl4}}{\text{volume}} = \dfrac{0.0437\,\text{mol}}{62.92 \times 10^{-3}\, \text{L}} = 0.694\,\text{M}$$ Since there is no dimer to start, \([\text{V2Cl8}]_{initial} = 0\,\text{M}\). Partial equilibrium equation: $$2 \,\text{VCl4} \rightleftharpoons \text{V2Cl8}$$ $$(0.694 - 2x_\text{eq})\quad \text{--->}\quad \quad x_\text{eq}$$

Now using the initial concentrations and the molality we calculated in step 1, we can calculate the moles of VCl4 and V2Cl8 at equilibrium based on their mole fraction in the solution $$\text{mole fraction}_{VCl4} = \dfrac{\text{moles}_{VCl4}}{\text{moles}_{VCl4} + \dfrac{1}{2}\text{moles}_{V2Cl8}}$$ Using the molality and moles of carbon tetrachloride, we can calculate the mole fraction $$\text{x}_{VCl4} = \dfrac{\text{molality} \times \text{moles}_{CCl4}}{\text{moles}_{VCl4} + \dfrac{1}{2}\text{moles}_{V2Cl8}}$$ $$0.2\, \text{mol/kg} × 0.65\, \text{mol} = \dfrac{0.0437\, \text{mol}-2x_\text{eq}+x_\text{eq}}{0.0437\, \text{mol}+0.5x_\text{eq}}$$ By solving this equation, we get: $$x_\text{eq} = 0.0235\, \text{M}$$ Now, we can find the \([\text{VCl4}]_{eq}\) and \([\text{V2Cl8}]_{eq}\). $$[\text{VCl4}]_{eq} = 0.694\, \text{M} - 2x_\text{eq} = 0.694\, \text{M} - 2(0.0235\, \text{M}) = 0.647\, \text{M}$$ $$[\text{V2Cl8}]_{eq} = x_\text{eq} = 0.0235\, \text{M}$$

Finally, we can calculate the equilibrium constant using the concentrations of VCl4 and V2Cl8 at equilibrium. $$K_{eq} = \dfrac{[\text{V2Cl8}]_{eq}}{[\text{VCl4}]_{eq}^2} = \dfrac{0.0235\, \text{M}}{(0.647\, \text{M})^2} = 5.62 \times 10^{-2}$$ Hence, the equilibrium constant for the dimerization of VCl4 at this temperature is \(5.62 \times 10^{-2}\).