1. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the vapor
ID: 1002326 • Letter: 1
Question
1. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the vapor pressures of pentane and hexane are 511 and 150 torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 36 mL hexane (density, 0.66 g/mL). What is the vapor pressure of the resulting solution? Ptotal = ___ torr 2. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the vapor pressures of pentane and hexane are 511 and 150 torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 36 mL hexane (density, 0.66 g/mL). What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? X V pen = ___Explanation / Answer
Q.1:
Mass of pentane in the solution = Vxd = 25 mL x 0.63 g/mL = 15.75 g
molar mass of pentane = 72.15 g/mol
Hence moles of pentane = mass/molar mass = 15.75 g / 72.15 g/mol = 0.2183 mol pentane
Mass of hexane in the solution = Vxd = 36 mL x 0.66 g/mL = 23.76 g
molar mass of hexane = 86.18 g/mol
Hence moles of hexane = mass/molar mass = 23.76 g / 86.18 g/mol = 0.2757 mol hexane
The vapor pressure of the resulting solution can be calculated from Raoult's law
Ptotal = Xpentane * P0pentane + Xhexane * P0hexane
=> Ptotal = (0.2183 mol / 0.2183 mol + 0.2757 mol) * 511 torr + (0.2757 mol / 0.2183 mol + 0.2757 mol) * 150 torr
=> Ptotal = 225.81 torr + 83.715 torr = 309.5 torr (answer)
Q.2: Partial pressure of pentane = 225.81 torr
Total pressure of the solution = 309.5 torr
For the vapor phase
Partial pressure of pentane = (moles fraction of pentane) x total pressure of the solution
=> 225.81 torr = XVpentane x 309.5 torr
=> XVpentane = 225.81 torr / 309.5 torr = 0.730 (answer)
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