Question

At 330 K the vapor pressure of pure n-pentane is 1.92 atm and the vapor pressure of pure n-octane is 0.07 atm. If 330K is the normal boiling point for a solution of these two substances, what will the mole fractions of each substance be in that solution

Answer 1

Answer: mole fractions are

For n-pentane = 0.965

For n-octane = 0.035

Explanation: pressure exerted by each gas is,

n-pentane = 1.92atm

n-octane = 0.07atm

Total pressure exerted = 1.92 + 0.07

= 1.99atm.

Recall that the partial pressure exerted by each gas is the product of its mole fraction and the total pressure, that is,

Pres. n-pentane = n x pressure(total)

1.92 = n x 1.99

n = 1.92/1.99 = 0.965 for n-pentane

For n-octane,

n = 1 - 0.965 = 0.035 for n-octane.