Question

At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepared by mixing 127.g of acetic acid and 141.g of methanol CH3OH. Calculate the partial pressure of acetic acid vapor above this solution. Round your answer to 3 significant digits.

Answer 1

The partial pressure of acetic acid is 73.5 torr

Step-by-step explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr

The partial pressure of acetic acid is 73.5 torr