What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg? (Assume dilute…

Strategy:\n> According to Raoult's law for non-volatile solutes, the relative lowering of vapour pressure is equal to the mole fraction of the solute ($\\frac{P^\circ - P}{P^\circ} = \chi_{\\text{solute}}$).\n\nStep 1: Identify given values\n- $P^\circ = 54.2\ \\text{mm Hg}$ (Pure water)\n- $P^\circ - P = \Delta P = 0.20\ \\text{mm Hg}$\n- Mass of water ($W$) = 100 g\n- Molar mass of water = 18 g/mol\n- Molar mass of glucose = 180 g/mol\n\nStep 2: Calculate mole fraction of glucose\n$\chi_{\\text{glucose}} = \\frac{\Delta P}{P^\circ} = \\frac{0.20}{54.2} \approx 0.00369$\n\nStep 3: Calculate moles of water ($n_w$)\n$n_w = \\frac{100}{18} \approx 5.556\\text{ mol}$\n\nStep 4: Solve for mass of glucose ($w_g$)\nFor a dilute solution: $\chi_g \approx \\frac{n_g}{n_w}$\n$\\frac{w_g / 180}{5.556} = 0.00369 \implies \\frac{w_g}{180} = 0.0205$\n$w_g = 180 \\times 0.0205 = 3.69\\text{ g}$\n\n$\boxed{\\text{Answer: (2)}}$