What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0 . 20 mm Hg ? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is 54 . 2 mm Hg at room temperature.Molar mass of glucose is 180 g mol - 1

Step-by-step solution

Sol. Using the formula P o - P s P o = Moles   of   solute Moles   of   solvent for dilute solution P o = Vapour pressure of pure water P s = Vapour pressure of solution 54 . 2 - 54 54 . 2 = w / 180 100 18 w = 0 . 2 × 1000 54 . 2 = 200 54 . 2 = 3 . 69