Answer:
The partial pressure of acetic acid is 73.5 torr
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
