Pentane, C5H12, and hexane, C6H14, combine to form an ideal solution. At 25⁰C the vapor pressure of pentane is 511 torr and that of hexane is 150 torr. In the vapor over a pentane-hexane solution at 25⁰C, the mole fraction of pentane is 0.15. What is the mole fraction of pentane in the solution?
Hmmmm.
.."...,the mole fraction of pentane is .15. What is the mole fraction of pentane?
What am I missing?
To find the mole fraction of pentane in the solution, we need to use Raoult's law, which states that the vapor pressure of a component in an ideal solution is proportional to its mole fraction in the solution.
Raoult's Law equation:
P_total = X_pentane * P_pentane + X_hexane * P_hexane
Where:
- P_total is the total vapor pressure of the solution
- X_pentane is the mole fraction of pentane
- P_pentane is the vapor pressure of pentane
- X_hexane is the mole fraction of hexane
- P_hexane is the vapor pressure of hexane
To solve for the mole fraction of pentane (X_pentane), rearrange the equation:
X_pentane = (P_total - X_hexane * P_hexane) / P_pentane
Given values:
P_pentane = 511 torr
P_hexane = 150 torr
X_pentane = 0.15
We need to find P_total to calculate X_pentane.
To find P_total, we can use Dalton's law of partial pressures, which states that the total pressure of a gas mixture is the sum of the partial pressures of its components.
P_total = P_pentane + P_hexane
Substituting the given values, we have:
P_total = 511 torr + 150 torr
P_total = 661 torr
Now, we can substitute the calculated values into the rearranged equation to find X_pentane:
X_pentane = (P_total - X_hexane * P_hexane) / P_pentane
X_pentane = (661 torr - 0.85 * 150 torr) / 511 torr
X_pentane = (661 torr - 127.5 torr) / 511 torr
X_pentane = 533.5 torr / 511 torr
X_pentane ≈ 1.044
Therefore, the mole fraction of pentane in the solution is approximately 1.044.