chemistry problem, vapor pressure.?
Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 12.9 torr at 35°C. The vapor pressure of pure ethanol at 35°C is 1.00 102 torr.
work would be GREATTTLLYYY appreciated. thanks a million 
Comments
moles ethanol = 1000 g/ 46.069 g/mol= 21.7
vapor pressure solution must be 1.00 x 10^2 - 12.9 = 87.1 torr
87.1 = 1.00 x 10^2 X
X = mole fraction water = 0.871
0.871 = 21.7 / 21.7 + moles glycol
18.9 + 0.871 moles glycol = 21.7
0.871 moles glycol = 2.8
moles glycol = 3.21
mass glycol = 3.21 x 62.068 g/mol= 199.2 g