A. A solution of hexane and heptane at 30 degrees C with hexane mole fraction 0.
ID: 797855 • Letter: A
Question
A. A solution of hexane and heptane at 30 degrees C with hexane mole fraction 0.305 has a vapor pressure of 95.0 torr and a vapor-phase hexane mole fraction of 0.55. Find the vapor pressures of pure hexane and heptane at 30 degrees C. State any approximations made.
B. At 25 degrees C and 1 atm, a solution of 72.061 g of H2O and 192.252 g of CH3OH has a volume of 307.09 cubic cm. In this solution the partial molar volume of water is 16. 488 cubic cm/mol. Find the partial molar volume of ethanol in this solution.
Explanation / Answer
Let, p1 be the vapor pressure of hexane
p2 be the vapor pressure of heptane.
By raoult's law,
95.0 = 0.305(p1) + 0.695(p2)
Vapor phase hexane mole fraction = 0.55
=> 0.305(p1)/95 = 0.55
=> p1 = 171.31 torr
=> 0.695(p2) = (1-0.55)95 = 42.75
=> p2 = 61.51 torr
Assumption : Hexane and heptane mixture is an ideal mixture obeying raoult's law
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Moles of water = 4 moles
Moles of methanol = 6 moles
Total volume = 307.09 cc
Volume of water = 4*16.488 = 65.952 cc
Volume of methanol = 307.09- 65.952 =241.138 cc
Partial molar volume of methanol = 241.138/6 =40.19 cubic cm/mol
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