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telo118 [61]
3 years ago
12

The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com

position of the solution? Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.
Chemistry
1 answer:
FrozenT [24]3 years ago
8 0
  • The mass percent of Pentane in solution is 16.49%
  • The mass percent of Hexane in solution is 83.51%

<u>Explanation</u>:

  • Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.
  • Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

  • From Raoult's law we know:  

Pp = xp \times Pp - vap = yp \times Pt                                       (1)

Ph = xh \times Ph - vap = yh \times Pt                                       (2)

  • Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) \times Ph - vap = (1 - yp) \times Pt                                   (3)

  • Substituting (1) into (3) we get:

(1-xp) \times Ph - vap = (1 - yp) \times xp \times Pp - vap / yp            (4)

  • Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap)      (5)

  • Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%  

Hexane mol% in solution: xh = 80.92%

  • Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol \times 72.15 g/mol

     = 13.766 g

mh = 0.8092 mol \times 86.18 g/mol

     = 69.737 g

  • Mass% of Pentane solution = 13.766/(13.766+69.737)

                                                       = 16.49%

  • Mass% of Hexane solution  = 83.51%
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