Molecules of benzoic acid (C6H5COOH) dimerise in 30 g of benzene. 'w' g of benzoic acid shows a depression in freezing…

Strategy:\n> Benzoic acid undergoes association (dimerization) in benzene. First, find the van't Hoff factor for the given percentage association, then use the depression formula to solve for the required mass $w$.\n\nStep 1: Calculate the van't Hoff factor ($i$)\nFor dimerization ($n=2$) with $\alpha = 0.80$:\n$i = 1 - \alpha \left(1 - \\frac{1}{n}\right) = 1 - 0.8 \left(1 - \\frac{1}{2}\right) = 1 - 0.4 = 0.6$\n\nStep 2: Calculate the required molality ($m$)\n$\Delta T_f = 2\\text{ K}$, $K_f = 5$.\n$2 = 0.6 \\times 5 \\times m \implies m = \\frac{2}{3}\\text{ mol/kg}$\n\nStep 3: Solve for mass } w \text{\nSolvent mass = 30 g = 0.03 kg.\n$\\text{Moles } (n) = m \\times 0.03 = \\frac{2}{3} \\times 0.03 = 0.02\\text{ mol}$\nMass of acid ($w$) = moles $\times$ molar mass = $0.02 \\times 122 = 2.44\\text{ g}$.\nRounding to nearest tenth: 2.4.\n\n$\boxed{\\text{Answer: (2)}}$