Benzene and toluene form nearly ideal solution. At 20°C,the vapour...
°C, the vapor pressure of pure benzene is 74 torr and the vapor pressure of pure toluene is 22 torr. When a solution containing 50.0 g of benzene and 50.0 g of toluene is prepared, the vapor pressure of the solution is 38 torr. Calculate the mole fraction of benzene in the solution.
To solve this problem, we can use Raoult's Law, which states that the partial pressure of a component in a solution is equal to the product of its mole fraction and its vapor pressure in its pure state.
Let's denote the mole fraction of benzene as x and the mole fraction of toluene as (1-x).
According to Raoult's Law, the partial pressure of benzene (Pbenzene) in the solution is given by:
Pbenzene = x * Pbenzene(pure)
Similarly, the partial pressure of toluene (Ptoluene) in the solution is given by:
Ptoluene = (1-x) * Ptoluene(pure)
Given that Pbenzene(pure) = 74 torr, Ptoluene(pure) = 22 torr, and the total vapor pressure of the solution (Ptotal) = 38 torr, we can set up the following equation:
Pbenzene + Ptoluene = Ptotal
Substituting the expressions for Pbenzene and Ptoluene:
x * Pbenzene(pure) + (1-x) * Ptoluene(pure) = Ptotal
x * 74 + (1-x) * 22 = 38
Simplifying the equation:
74x + 22 - 22x = 38
52x = 16
x = 16/52
x = 0.3077
Therefore, the mole fraction of benzene in the solution is approximately 0.3077.
Benzene and toluene form nearly ideal solution. At 20°C,the vapour...
According to Raoult's law, Partial vapour pressure = mole fraction × pressure by its pure solution.
Thus, you have to just calculate value of mole fraction of benzene and multiply it with pressure of Benzene which is given question. You will get answer as 50torr.
If you want full solution then reply and ask.