What is the vapor pressure at 23°C of a solution of 1.20 g of naphthalene, C10H8, in 25.6 g of benzene,
C6H6? The vapor pressure of pure benzene at 23°C is 86.0 mmHg; the vapor pressure of naphthalene
can be neglected. Calculate the vapor-pressure lowering of the solution.
The Raoult's law says that the vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent. The vapor pressure lowering is directly proportional to the mole fraction of the solute. In mathematical terms that means:
ΔP = (1)
ΔP is the vapor-pressure lowering of the solution is the molar fraction of the solute and is the vapor pressure of the pure solvent (86,0 mmHg)
The moles of naphthalene are:
9,36x10⁻³ moles solute
The moles of benzene are:
0,328 moles solvent
Molar fraction of solute is:
0,0278
Replacing in (1)
ΔP = 2,39 mm Hg
Therefore, the vapor-pressure lowering of the solution is 2,39 mmHg
Why did the solution seek therapy? Because it had a low self-esteem due to its vapor pressure being lowered.
To calculate the vapor pressure of the solution and the vapor-pressure lowering, we can use Raoult's Law.
Raoult's Law states that the vapor pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in its pure state.
First, let's calculate the mole fraction of benzene in the solution:
Moles of benzene = (mass of benzene) / (molar mass of benzene)
Moles of benzene = 25.6 g / 78.11 g/mol (molar mass of benzene)
Moles of benzene = 0.328 mol
Moles of naphthalene = (mass of naphthalene) / (molar mass of naphthalene)
Moles of naphthalene = 1.20 g / 128.17 g/mol (molar mass of naphthalene)
Moles of naphthalene = 0.00936 mol
Total moles of both components = Moles of benzene + Moles of naphthalene
Total moles of both components = 0.328 mol + 0.00936 mol
Total moles of both components = 0.337 mol
Next, let's calculate the mole fraction of benzene:
Mole fraction of benzene = (moles of benzene) / (total moles of both components)
Mole fraction of benzene = 0.328 mol / 0.337 mol
Mole fraction of benzene = 0.972
Now, we can use Raoult's Law to calculate the vapor pressure of the solution:
Vapor pressure of the solution = (mole fraction of benzene) * (vapor pressure of pure benzene)
Vapor pressure of the solution = 0.972 * 86.0 mmHg (vapor pressure of pure benzene)
Vapor pressure of the solution = 83.67 mmHg
Finally, let's calculate the vapor-pressure lowering of the solution:
Vapor-pressure lowering = (vapor pressure of pure benzene) - (vapor pressure of the solution)
Vapor-pressure lowering = 86.0 mmHg - 83.67 mmHg
Vapor-pressure lowering = 2.33 mmHg
Therefore, the vapor-pressure lowering of the solution is 2.33 mmHg.
To calculate the vapor-pressure lowering of the solution, we need to use Raoult's law. Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
First, let's calculate the mole fraction of benzene in the solution:
Moles of benzene = mass of benzene / molar mass of benzene
Moles of benzene = 25.6 g / 78.11 g/mol (molar mass of benzene)
Next, let's calculate the mole fraction of naphthalene in the solution:
Moles of naphthalene = mass of naphthalene / molar mass of naphthalene
Moles of naphthalene = 1.20 g / 128.17 g/mol (molar mass of naphthalene)
The mole fraction of benzene is then given by:
Mole fraction of benzene = Moles of benzene / (Moles of benzene + Moles of naphthalene)
Finally, we can use Raoult's law to calculate the vapor pressure of the solution:
Vapor pressure of solution = Vapor pressure of pure benzene * Mole fraction of benzene
Vapor-pressure lowering of the solution is then given by:
Vapor-pressure lowering = Vapor pressure of pure benzene - Vapor pressure of solution
Substituting the given values, we can now calculate the vapor-pressure lowering.