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 $180 g {mol}^{- 1}$

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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

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