Question

A 300 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

Answer 1

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, X = 16.67 moles water/17.07 moles total = 0.977 and the new pressure = 0.977 x 740 mm Hg = 723 mm Hg.