A mixture is prepared by mixing 100.0 grams of ethanol (C2H6O) with 30.0 grams of acetone (C3H6O). At 25°C the vapor pressures of pure ethanol and acetone are 59 Torr and 231 Torr respectively. What will be the total vapor pressure over the mixture at this temperature? a. 212 Torr
b. 62 Torr
c. 92 Torr
d. 88 Torr
e. 46 Torr
Explanation: Total vapor pressure is equal to the sum of partial pressures of each components of the mixture.
How do we calculate partial pressure when pure vapor pressure is given?
Pa= XaPºa, where Pa is the new vapor pressure of A, Xa is the mole fraction of A = moles of A/total moles and Pºa is the pure vapor pressure of A.
Let's begin by calculating moles and then mole fraction of each component.
100.0 grams of ethanol/ (46.07 g/mol) = 2.17 mol of ethanol
30.0 grams of acetone / 58.08 g/mol = 0.517mol of acetone
Moles fraction of ethanol = 2.17 mo/ (2.17 mol +0.517mol) = 0.808
Mole fraction of acetone = 1-0.808 = 0.192 (all mole fractions add up to 1)
Pethanol = 0.808(59 Torr ) = 47.67 torr
Pacetone = 0.192(231 Torr) = 44.35 torr
Ptotal = Pethanol + Pacetone = 47.67 torr+44.35 torr = 92.022 torr