Asked by jonny
What is the vapor pressure (in mmHg) of a solution containing 16.0 g of (3 marks) glucose (C6H12O6) in 80.0 g of methanol (CH3OH)? The vapor pressure of pure methanol at 27 °C is 140 mm Hg. If the experimentally determined vapor pressure is 128 mm Hg, account for the difference between the two values. (3 significant figures).
Answers
Answered by
DrBob222
Please don't change screen names.
mols glucose = grams/molar mass
mols methanol = grams/molar mass
Total mols = mols glucose + mols methanol
mole fraction methanol = nmethanol/total mols.
psoln = X<sub>methanol</sub>*P<sup>o</sup><sub>methanol</sub>
For the difference, what about interactions between solvent molecules. What about interactions between solvent and solute?
mols glucose = grams/molar mass
mols methanol = grams/molar mass
Total mols = mols glucose + mols methanol
mole fraction methanol = nmethanol/total mols.
psoln = X<sub>methanol</sub>*P<sup>o</sup><sub>methanol</sub>
For the difference, what about interactions between solvent molecules. What about interactions between solvent and solute?
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