Question

At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the lowering of the vapor pressure of pentane in this solution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution.

a. 115 torr
b. 138 torr
c. 161 torr
d. 187 torr
e. 206 tor

Answer 1

The lowering of the vapor pressure of pentane in the solution at 25°C, calculated using Raoult's Law, is approximately 161 torr, which is option (c) from the multiple-choice list.

Step-by-step explanation:

To calculate the lowering of the vapor pressure of pentane in the solution, we can use Raoult's Law, which states that the partial vapor pressure of a component in an ideal solution is equal to the mole fraction of that component in the solution multiplied by the vapor pressure of that component as a pure substance.

The mole fraction of pentane (χC5H12) is calculated as the moles of pentane divided by the total moles of both components:

χC5H12 = moles of C5H12 / (moles of C5H12 + moles of C5H10) = 0.450 / (0.450 + 0.250) = 0.643

The lowering of the vapor pressure of pentane (ΔPC5H12) is then:
Pure vapor pressure of pentane - (χC5H12 × Vapor pressure of pentane)

ΔPC5H12 = 451 torr - (0.643 × 451 torr) = 451 torr - 290.093 torr = 160.907 torr

The closest answer to the calculated value is 161 torr (± 1 torr due to rounding), which corresponds to the multiple-choice option (c).

\