the molar volume of liquid benzene (density=0.877gm/ml) increased by a factor of 2750 as it vaporises at 20 degree celcius and that of liquid toluene (density=0.867gm/ml)increases by a factor of 7720 at 20 degree celcius . a solution of benzene and toluene at 20 degee celcius has a vapour pressure of 46 torr . find the mole fraction 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 at 20 degrees Celsius, we can use Raoult's Law and the information provided about the molar volumes and vapor pressures. Let's break this down step by step.

Understanding the Problem

We know the following:

• The density of liquid benzene is 0.877 g/mL.
• The density of liquid toluene is 0.867 g/mL.
• The molar volume of benzene increases by a factor of 2750 upon vaporization.
• The molar volume of toluene increases by a factor of 7720 upon vaporization.
• The total vapor pressure of the solution at 20 degrees Celsius is 46 torr.

Calculating Molar Masses and Molar Volumes

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

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

Next, we can calculate the molar volumes of the liquids:

• Molar volume of benzene = (molar mass / density) = (78.11 g/mol) / (0.877 g/mL) ≈ 89.0 mL/mol
• Molar volume of toluene = (molar mass / density) = (92.14 g/mol) / (0.867 g/mL) ≈ 106.0 mL/mol

Calculating Vapor Pressures

Using the factors by which the molar volumes increase upon vaporization, we can find the molar volumes of the gases:

• Vapor volume of benzene = 2750 × 89.0 mL/mol ≈ 244750 mL/mol
• Vapor volume of toluene = 7720 × 106.0 mL/mol ≈ 819520 mL/mol

Applying Raoult's Law

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

• Partial pressure of benzene (P_B) = X_B × P_B0
• Partial pressure of toluene (P_T) = X_T × P_T0

Where:

• X_B and X_T are the mole fractions of benzene and toluene, respectively.
• P_B0 and P_T0 are the vapor pressures of pure benzene and toluene at 20 degrees Celsius.

Finding Vapor Pressures of Pure Components

To find the vapor pressures of pure benzene and toluene, we can use the ideal gas law and the molar volumes we calculated:

• P_B0 = (RT) / V_B = (0.0821 L·atm/(K·mol) × 293 K) / (244.75 L/mol) ≈ 0.101 atm ≈ 76 torr
• P_T0 = (RT) / V_T = (0.0821 L·atm/(K·mol) × 293 K) / (819.52 L/mol) ≈ 0.024 atm ≈ 18 torr

Setting Up the Equation

Now we can set up the equation for the total vapor pressure:

P_total = P_B + P_T = 46 torr

Substituting the expressions for partial pressures:

X_B × 76 torr + X_T × 18 torr = 46 torr

Since X_B + X_T = 1, we can express X_T as (1 - X_B):

X_B × 76 + (1 - X_B) × 18 = 46

Solving for Mole Fraction

Now, let's solve for X_B:

76X_B + 18 - 18X_B = 46

58X_B = 28

X_B = 28 / 58 ≈ 0.483

Final Result

The mole fraction of benzene in the vapor above the solution is approximately 0.483. This means that about 48.3% of the vapor consists of benzene molecules at 20 degrees Celsius.