Pentane (C5H12) and hexane (C6H14) form an ideal solution. The vapour pressure of pentane at 25oC is 511 mmHg and that for hexane is 150 mmHg. A solution is prepared by mixing pentane (density = 0.63 g/mL) and hexane (density = 0.66 g/mL). The mole fraction of hexane in the vapour phase in equilibrium with this solution is 0.4. Calculate the volume of hexane that would be present in 1.0 L of this solution.
To solve this problem, we can use the concept of Raoult's Law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.
First, let's find the mole fraction of pentane and hexane in the liquid phase.
Given:
Vapor pressure of pentane (P₁) = 511 mmHg
Vapor pressure of hexane (P₂) = 150 mmHg
Using Raoult's Law:
P₁ = X₁ * P°₁
P₂ = X₂ * P°₂
Where:
X₁ is the mole fraction of pentane
X₂ is the mole fraction of hexane
P°₁ is the vapor pressure of pure pentane
P°₂ is the vapor pressure of pure hexane
Since the mole fraction of hexane in the vapor phase is given as 0.4, we can use this information to find X₂ in the liquid phase using the equation:
X₂ = (X₂ Vapor * P₂) / (X₁ Vapor * P₁ + X₂ Vapor * P₂)
Rearranging the equation, we get:
X₂ Liquid = (X₂ Vapor * P₂) / (P₁ - (X₁ Vapor * (P₁ - P₂)))
Now we need to find X₁ Vapor, the mole fraction of pentane in the vapor phase. We can use the equation:
X₁ Vapor = (P₁ - X₂ Vapor * P₂) / (P₁ - P₂)
Substituting the given values:
X₂ Vapor = 0.4
P₁ = 511 mmHg
P₂ = 150 mmHg
Calculating X₁ Vapor:
X₁ Vapor = (511 - 0.4 * 150) / (511 - 150)
= 0.706
Now we can substitute the values of X₂ Vapor and X₁ Vapor into the equation for X₂ Liquid to find X₂ Liquid:
X₂ Liquid = (0.4 * 150) / (511 - (0.706 * (511 - 150)))
= 0.315
Now we can calculate the moles of pentane and hexane in the solution:
moles of pentane = X₁ Liquid * volume * density of pentane
moles of hexane = X₂ Liquid * volume * density of hexane
Since the total volume of the solution is given as 1.0 L, we can substitute the given values and solve for the volume of hexane:
1.0 * 0.63 * moles of pentane + 1.0 * 0.66 * moles of hexane = 1.0 * 0.66 * volume of hexane
Simplifying the equation:
0.63 * moles of pentane + 0.66 * moles of hexane = 0.66 * volume of hexane
Substituting the calculated values of X₁ Liquid and X₂ Liquid:
0.63 * (X₁ Liquid * 1.0 * 0.63) + 0.66 * (X₂ Liquid * 1.0 * 0.66) = 0.66 * volume of hexane
Solving the equation will give us the volume of hexane present in 1.0 L of the solution.