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.

The vapor pressure of the solution is 105.4 torr

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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

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