A300 gram sample of pure water exerts a vapor pressure of 740 millimeters of mercury on the walls of its container. if 0.4 00 moles of glucose (c6h12o6) is added to the water, what will be the vapor pressure of the resulting solution? 723 mm hg 738 mm hg 739 mm hg 743 mm hg

723 mm Hg

Explanation:The equation,   P solution = X solvent P solvent, can be used to determine vapor pressure lowering. 300 grams of water is 300g x 1 mole/18.0g = 16.67 moles water. Thus,   Imported Asset    = 16.67 moles water/17.07 moles total = 0.977 and the new pressure = 0.977 x 740 mm Hg = 723 mm Hg.

738 mm Hg

Explanation:

The equation, P solution = X solvent P° solvent,  can be used to determine vapor pressure lowering. 300 grams of water is 300g x 1 mole/18.0g = 16.67 moles water. Thus,   = 16.67 moles water/16.94 moles total = 0.984 and the new pressure = 0.984 x 750 mm Hg = 738 mm Hg.

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