Question

What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)

Answer 1

Answer:

Step 1 of 6

(a)

The mass of benzene is  , so calculate the moles of benzene as follows:



The mass of toluene is, so calculate the moles of toluene as follows:



Now, calculate the mole fraction as follows:





Therefore, the mole fraction of benzene and toluene is  and  respectively.

Step 2 of 6

(b)

The formula to calculate the partial pressure is as follows:



Here,  is the partial pressure of benzene,  is the vapour pressure of pure benzene and  is the mole fraction of benzene.

Vapour pressure of pure benzene at  is.

Substitute the values in the equation as follows:



Therefore, the partial pressure is  .

Step 3 of 6

(c)

Vapor pressure of the solution at 1 atm is  .

When the total pressure of the vapour pressure of the mixture is  at a temperature, then, the solution boils. It corresponds to the boiling point of the solution.

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is less than the atmospheric pressure, the solution will not boil at  .

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is greater than the atmospheric pressure, the solution will boil at  .

Therefore, the boiling point of the solution is  .

Step 4 of 6

(d)

Mole fraction of benzene at  is calculated as follows:



Mole fraction of toluene at  is calculated as follows:



Therefore, the mole fractions of benzene and toluene are  and  respectively.

Step 5 of 6

(e)

Vapor pressure of benzene at  is  .

Partial pressure of benzene is calculated as follows:



Vapor pressure of toluene at  is  .

Partial pressure of toluene is calculated as follows:



Step 6 of 6

Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Explanation:

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Answer 2

Total mol = (3.9 g * 1 mol C6H6 / 78.11 g = 0.0499 mol) + (4.6 g * 1 mol C7H8 / 92.14 = 0.0499 mol) =0.0798 mol.
Mole fraction benzene = 0.0499 / 0.0798 = 0.5.
Mole fraction toluene = 0.0499 / 0.0798 = 0.5