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Chemistry Homework Help => General Chemistry => Topic started by: dtruong09 on June 19, 2010, 06:58:58 PM

Title: Colligative properties
Post by: dtruong09 on June 19, 2010, 06:58:58 PM
Calculate the mass of propylene glycol  that must be added to 0.320g  of water to reduce the vapor pressure by 2.87 torr  at 40C  (PH2O at 40C = 55.3 torr ).
Title: Re: Colligative properties
Post by: uma on June 21, 2010, 10:20:59 AM
 we know that
P(solution)= molefraction of H2O X P (pure water )-------eq1
now P(solution)= 55.3 torr - 2.87 torr 
P (pure water ) = 55.3 tor
Calculate mole fraction of H2O from eq1
Now mole fraction H2O = nH2O/ (nH2O + n propyleneglycol)  ---------eq2
nH2O  = mass of water / molar mass = 0.320 / 18
plug in all known values in eq 2 to get the mole  of propylene glycol
mass of propylene glycol = moles X  molar mass of propylene glycol