Question

Calculate the vapor pressure of a solution prepared by dissolving 5 g of water in 78 g of ethanol (CH3CH2OH) at 37 C. The vapor pressures of ethanol and water at this temperature are 115 torr (ethanol) and 47 torr (water), respectively.

Answer 1

The vapor pressure of the solution is 105.4 torr

Step-by-step explanation:

Step 1: Data given

MAss of water = 5.0 grams

Mass of ethanol = 78.0 grams

Temperature = 37 °C = 310 K

Molar mass water = 18.02 g/mol

Molar mass ethanol = 46.07 g/mol

The vapor pressures of ethanol = 115 torr

The vapor pressures of water at this temperature = 47 torr

Step 2: Calculate moles

Moles = mass / molar mass

Moles water = 5.0 grams / 18.02 g/mol

Moles water = 0.277 moles

Moles ethanol = 78.0 grams / 46.07 g/mol

Moles ethanol = 1.693 moles

Step 3: Calculate total moles

Total moles = 0.277 moles + 1.693 moles

Total moles = 1.97 moles

Step 4: Calculate mol fraction

Mol fraction = moles / total moles

Mol fraction H2O = 0.277 moles / 1.97 moles

Mol fraction H2O = 0.141

Mol fraction ethanol = 1.693 moles/1.97 moles

Mol fraction ethanol = 0.859

Step 5: Calculate the vapor pressure of the solution

Vapor pressure solution = pH2O * XH2O + pethanol * Xethanol

⇒with pH2O = the vapor pressure of H2O = 47 torr

⇒with XH2O = the mol fraction of H2O = 0.141

⇒with pEthanol = vapor pressure of ethanol = 115 torr

⇒with Xethanol = mol fraction of ethanol = 0.859

Vapor pressure solution = 47*0.141 + 115*0.859

Vapor pressure solution = 105.4 torr

The vapor pressure of the solution is 105.4 torr

Answer 2

Pv sln = 110.902 torr

Step-by-step explanation:

• Pv sln = Pv H2O*XH2O + Pvethanol*Xethanol

∴ m sln = 5g H2O + 78g ethanol = 83 g

⇒ X H2O = m H2O/m sln = 5 / 83 = 0.0602

⇒ X ethanol = 78 / 83 = 0.9396

∴ Pv H2O = 47 torr

∴ Pv ethanol = 115 torr

⇒ Pv sln = ((47 torr)*(0.0602)) + ((115 torr)*(0.9396))

⇒ Pv sln = 110.902 torr