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A mixture is prepared by mixing 100.0 grams of ethanol (C2H6O) with 30.0 grams of acetone (C3H6O). A

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


Answer: C

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