Asked by Vanessa
A closed bottle contains a solution of 5.0% by mass acetone [(CH3)2CO] in water. Calculate the vapour pressure (in kPa) of this solution at 25oC. The vapour pressure of pure water at 25oC is 3.17 kPa and that of pure acetone is 30.7 kPa.
Answers
Answered by
DrBob222
If you assume this is an ideal solution. Take 100 g solution.
mass acetone = 5g
mols acetone = 5/molar mass = ?
mass H2O = 95 g
mols H2O = 95/molar mass H2O = ?
total mols = mols acetone = mols H2O = ?
Xacetone = mols fraction acetone = nacetone/total mols
xH2O = nH2O/total mols.
pacetone = Xacetone*Poacetone
pH2O = XH2O*PoH2O
Total P = pacetone + pH2O
mass acetone = 5g
mols acetone = 5/molar mass = ?
mass H2O = 95 g
mols H2O = 95/molar mass H2O = ?
total mols = mols acetone = mols H2O = ?
Xacetone = mols fraction acetone = nacetone/total mols
xH2O = nH2O/total mols.
pacetone = Xacetone*Poacetone
pH2O = XH2O*PoH2O
Total P = pacetone + pH2O
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