Asked by Kris
How would you prepare 500.0 of an aqueous solution of glycerol with a vapor pressure of 24.8mm Hg at 26 degree Celsius (vapor pressure of pure water=25.21mm Hg)? assume the density of the solution to be 1.00 g/mL.
Answers
Answered by
DrBob222
500.0 what? I'll assume you meant 500.0 mL.
Psoln = XH2O*P<sup>o</sup>
Substitute and solve for XH2O = mole fraction H2O to produce vapor pressure of 24.80 mm.
If we let x = grams glycerol, then 500-x = grams H2O
mols glycerol = x/92.09
mols H2O = (500-x)/18.015
Plug into mole fraction H2O
nH2O/(nH2O + nglycerol) = XH2O and solve for g glycerol and g H2O.
Psoln = XH2O*P<sup>o</sup>
Substitute and solve for XH2O = mole fraction H2O to produce vapor pressure of 24.80 mm.
If we let x = grams glycerol, then 500-x = grams H2O
mols glycerol = x/92.09
mols H2O = (500-x)/18.015
Plug into mole fraction H2O
nH2O/(nH2O + nglycerol) = XH2O and solve for g glycerol and g H2O.
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