A solution is prepared by dissolving 181.2 g of glucose, C6H12O6, with 319.0 g of water at 100°C. By how much lower (in mm Hg) will be the vapor pressure of water above this solution be compared with pure water? The solute is a non-volatile compound.
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Well, I reasoned that
Change in P = (mole fraction of the Solute)(Pressure of the Pure solvent)
I'm given what I need to get the mole fraction of the Solute but I'm not given the pressure of the pure solvent so I can't solve for the change in pressure which is what the question is asking for.
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Topic: Vapor Pressure Above Water (Read 1654 times)