The molar volume of liquid benzene(Density=0.877) increases by a factor of 2750 as it vaporizes at 20 degrees celsius and that of liquid toluene(density=0.867) increases by a factor of 7720 at 20 degrees celcius.Solution of Benzene and toluene has a vapor pressure of 46 torr.

Find the molefraction of Benzene in the vapour above the solution.

To find the mole fraction of benzene in the vapor above a solution of benzene and toluene, we can use Raoult's Law, which relates the vapor pressure of each component in a solution to its mole fraction. Let's break down the problem step by step.

Understanding the Given Data

We have the following information:

• Density of Benzene: 0.877 g/mL
• Density of Toluene: 0.867 g/mL
• Vapor Pressure of the Solution: 46 torr
• Vapor Volume Increase Factor for Benzene: 2750
• Vapor Volume Increase Factor for Toluene: 7720

Calculating the Molar Masses

First, we need to determine the molar masses of benzene and toluene:

• Benzene (C₆H₆): 78.11 g/mol
• Toluene (C₇H₈): 92.14 g/mol

Finding the Molar Volumes

Next, we can calculate the molar volumes of benzene and toluene in their liquid states using their densities:

• Molar Volume of Benzene: V_B = (Molar Mass of Benzene) / (Density of Benzene) = 78.11 g/mol / 0.877 g/mL ≈ 89.00 mL/mol
• Molar Volume of Toluene: V_T = (Molar Mass of Toluene) / (Density of Toluene) = 92.14 g/mol / 0.867 g/mL ≈ 106.00 mL/mol

Calculating the Molar Volumes of Vapor

Now, we can find the molar volumes of the vapors using the given volume increase factors:

• Molar Volume of Vaporized Benzene: V_B,vapor = V_B × 2750 ≈ 89.00 mL/mol × 2750 ≈ 245,750 mL/mol
• Molar Volume of Vaporized Toluene: V_T,vapor = V_T × 7720 ≈ 106.00 mL/mol × 7720 ≈ 817,520 mL/mol

Applying Raoult's Law

According to Raoult's Law, the partial vapor pressure of each component in the solution can be expressed as:

• Partial Pressure of Benzene: P_B = X_B × P_B,0
• Partial Pressure of Toluene: P_T = X_T × P_T,0

Where:

• X_B: Mole fraction of benzene
• X_T: Mole fraction of toluene
• P_B,0: Vapor pressure of pure benzene
• P_T,0: Vapor pressure of pure toluene

Finding the Total Vapor Pressure

The total vapor pressure of the solution is given as 46 torr. We can express this as:

P_total = P_B + P_T

Substituting in the expressions from Raoult's Law:

P_total = X_B × P_B,0 + X_T × P_T,0

Calculating Mole Fractions

Since the mole fractions must add up to 1:

X_T = 1 - X_B

Substituting this into the total vapor pressure equation gives us:

46 = X_B × P_B,0 + (1 - X_B) × P_T,0

Finding Vapor Pressures of Pure Components

To proceed, we need the vapor pressures of pure benzene and toluene at 20 degrees Celsius. These values are typically found in tables:

• P_B,0 (Benzene): Approximately 75 torr
• P_T,0 (Toluene): Approximately 22 torr

Substituting Values

Now we can substitute these values into our equation:

46 = X_B × 75 + (1 - X_B) × 22

Expanding this gives:

46 = 75X_B + 22 - 22X_B

Combining like terms results in:

46 = 53X_B + 22

Solving for X_B:

24 = 53X_B

X_B = 24 / 53 ≈ 0.4528

Final Result

The mole fraction of benzene in the vapor above the solution is