A 300 gram sample of pure water exerts a vapor pressure of 750 millimeters of mercury on the walls of its container. If 0.27 moles of glucose (C6H12O6) is added to the water, what will be the vapor pressure of the resulting solution?

0.984 mm Hg

12.1 mm Hg

738 mm Hg

748 mm Hg

Question

A 300 gram sample of pure water exerts a vapor pressure of 750 millimeters of mercury on the walls of its container. If 0.27 moles of glucose (C6H12O6) is added to the water, what will be the vapor pressure of the resulting solution?
		
0.984 mm Hg
		
12.1 mm Hg
		
738 mm Hg
		
748 mm Hg

nvidiaintely · Answer

@sshayer @sweetburger @mathstudent55

nvidiaintely · Answer

@AloneS

nvidiaintely · Answer

@Nnesha

kg1975 · Answer

Psolution is the vapor pressure of the solution
Χsolvent is mole fraction of the solvent
P0solvent is the vapor pressure of the pure solvent

mole fraction = moles of solute/total solution moles

moles glucose = .27
finding moles water = 300/18 = 16.67
total moles solution = 16.67 + .27 = 16.93

mole fraction = .27/(16.93) = 0.0159

Rauolt's Law: Psolution = (Χsolvent)(P0solvent) = (0.0159)(1 atm)
i think 1 atm = 750 mmHg

Psolution = 0.984

kg1975 · Answer

believe me or not :T like i care 

but a few people aint gonna stop me from helping no matter what u think

nvidiaintely · Answer

The answer is 738 mmHg.

kg1975 · Answer

sure.. if u knew why did u ask even tho ur wrong

quickstudent · Answer

@kg1975 .............he simply submitted the quiz, and his teacher said that that was the real answer.