Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point…

Strategy:\n> Association leads to a decrease in the number of particles ($i < 1$). Calculate the actual molality from $\Delta T_f$, compare it to the expected molality to find $i$, and solve for the degree of association $\alpha$.\n\nStep 1: Calculate expected molality ($m_{exp}$)\n- Mass of A = 0.7 g, Molar mass = 93 g/mol.\n- Mass of water = 42 g = 0.042 kg.\n$m_{exp} = \\frac{0.7/93}{0.042} \approx 0.1793\\text{ mol/kg}$\n\nStep 2: Calculate observed molality ($m_{obs}$)\n$\Delta T_f = 0.2\\text{ K}$, $K_f = 1.86$.\n$m_{obs} = \\frac{0.2}{1.86} \approx 0.1075\\text{ mol/kg}$\n\nStep 3: Determine van't Hoff factor ($i$)\n$i = \\frac{m_{obs}}{m_{exp}} = \\frac{0.1075}{0.1793} \approx 0.6$\n\nStep 4: Solve for association\nAssuming dimerization ($2A \o A_2$): $i = 1 - \\frac{\alpha}{2}$.\n$0.6 = 1 - \\frac{\alpha}{2} \implies \alpha = 0.8 = 80\%$\n\n$\boxed{\\text{Answer: (4)}}$