Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 months ago

chemistry help?

"Cowboy coffee" is prepared by boiling coarse coffee grounds in water. If 12.5 g sucrose (C12H22O11) is added to 225 g cowboy coffee as it is being prepared, by how much is the vapor pressure above the solution reduced? Assume the solvent, the cowboy coffee, has the same properties as water. The vapor pressure of the solvent at 100 °C is 760.0 Torr.

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  • 3 months ago
    Favorite Answer

    Moles sucrose = 12.5 g / 342.3 g/mol = 0.0365 mol

    moles H2O = 225 g / 18.01 g/mol =12.5 mol water

    mole fraction water = (12.5 / (12.5+0.0365) = 0.997

    Vapor pressure solution = 760.0 torr X 0.997 = 757.8 torr

    So, the vapor pressure is reduced by 2.2 torr

  • 3 months ago

    use stoikieomettry

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