How many grams of glucose, C6H12O6, should be dissolved in 0.5 kg of water at 25°C to reduce the vap — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Reducing the vapour pressure by 1.0% means (P^o - P) / P^o = X_glucose = 0.01. Moles of water N = 500 / 18 = 27.78 mol. Using the exact formula for mole fraction: n / (n + N) = 0.01 => n / (n + 27.78) = 0.01 => n = 0.01n + 0.2778 => 0.99n = 0.2778 => n ≈ 0.2806 mol. Mass of glucose = n * molar mass of glucose = 0.2806 * 180 ≈ 50.5 g.