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Mass and Vapor Pressure

Submitted by sswa0322 on Wed, 12/02/2009 - 22:45

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 13.1 torr at 35°C. The vapor pressure of pure ethanol at 35°C is 1.00  102 torr.

relative lowering of vapor pressure is given
mole fraction of solute (ethylene glycol)= lowering in vapor pressure /vapor pressure of pure solvent

mole fraction of ethylene glycol = moles of ethylene glycol / total sum of moles (ethanol + ethylene glycol)
this you can use to take out the moles of  ethylene glycol
moles of ethanol = mass of ethanol / molar mass
now mass of ethylene glycol =  number of moles X molar mass of ethylene glycol