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27 December, 16:14

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solution is allowed to reach 90.0°C. The vapor pressure of pure water at this temperature is 528.8 mm Hg. The vapor pressure of the solution is 423.0 mm Hg. How many kg of water was present?

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  1. 27 December, 17:45
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    0.144 kg of water

    Explanation:

    From Raoult's law,

    Mole fraction of solvent = vapor pressure of solution : vapor pressure of solvent = 423 mmHg : 528.8 mmHg = 0.8

    Let the moles of solvent (water) be y

    Moles of solute (C3H8O3) = 2 mole

    Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

    Mole fraction of solvent = moles of solvent/total moles of solution

    0.8 = y / (y + 2)

    y = 0.8 (y + 2)

    y = 0.8y + 1.6

    y - 0.8y = 1.6

    0.2y = 1.6

    y = 1.6/0.2 = 8

    Moles of solvent (water) = 8 mol

    Mass of water = moles of water * MW = 8 mol * 18 g/mol = 144 g = 144/1000 = 0.144 kg
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