At 20degree C, the vapor pressures of pure 2-butanone*, 1-butanol*, and water ar

Question

At 20degree C, the vapor pressures of pure 2-butanone*, 1-butanol*, and water are 71.2,4.2, and 17.5 torr, respectively (condensed Lewis structures of 2-butanone and 1-butanol are shown Using Raoult's law, predict the vapor pressure of a solution made by mixing 0.73 moles of 2-butanone with 0.71 moles of water. A solution with the same predicted vapor pressure as solution is made by mixing 2-butanone with 1-butanol. Calculate the mole fraction of 2-butanone in solution. Which solution, (a) or (b), is more likely to obey Raoult's law

Explanation / Answer

Po (2-butanone) = 71.2 torr

Po (1-butanol) = 4.2 torr

Po (water) = 17.5 torr

(a). Moles of 2-butanone = 0.73

Moles of water = 0.77

Mole fraction of 2-butanone = 0.73 / (0.73 + 0.77)

= 0.486

Mole fraction of water = 0.77 / (0.73 + 0.77)

= 0.514

Psol = 0.486*71.2 + 0.514*17.5

Psol = 43.6 torr

(b). Let the moole fraction of 2-butanone be x.

Psol = x*Po (2-butanone) + (1 - x)*Po (1-butanol)

43.6 = x*71.2 + (1 - x)*4.2

43.6 = x*71.2 + 4.2 - 4.2*x

43.6 = 67*x + 4.2

x = 0.588

(c). Solution a is more likely to obey raoult's law.

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