Question

Pentane and hexane form an ideal solution. At 25degrees the vapour pressure of pentane and hexane are 511 and 150.0 torr, respectively. A solution is prepared by mixing 25ml pentane (density, 0.63g/ml) with 45 ml hexane (density, 0.66g/ml)

a) What is the vapour pressure of the resulting solution

b) What is the composition by mole fraction of pentane in the resulting vapour that is in equilibrium with this solution?

c) Is it reasonable to assume that this is an ideal solution? Explain.

Answer 1

a) vapour pressure of pentane = 511*(25*0.63)/72

=111.781 torr

vapour pressure of hexane = 150*45*0.66/86 (sonce molar mass is 86)

=

a) moles of pentane in the solution = 0.63*25/72

=0.21875 moles

moled pf hexane in the solution = 45*0.66/86

=0.3453 moles

so total vapour pressure = (0.21875/(0.28175+0.3454)) * 511 + (0.3454/(0.28175+0.3454))*150

=260.849 torr

b) mole fraction is proportional to pressure.so,

mole fraction of pentane = pressure of pentane/(total pressure)

=(0.21875/(0.28175+0.3454)) * 511/260.849

=0.683

c)yes the solutino is idea since pentane and hexane do not react.