a.)What is the partial vapor pressure of a benzene above solution obtained by mixing 44.7 g benzene, C6H6, and 77.3 g toluene, C6H5CH3, at 25 ∘C? At 25 ∘C the vapor pressure of C6H6 = 95.1 mmHg; the vapor pressure of C6H5CH3 = 28.4 mmHg.
b.)What is the partial vapor pressure of a toluene?
c.)What is the total vapor pressure of a solution?
a) To calculate the partial vapor pressure of benzene, we need to determine the mole fraction of benzene in the solution using the given masses of benzene and toluene.
Molar mass of benzene (C6H6) = 78.11 g/mol
Molar mass of toluene (C6H5CH3) = 92.14 g/mol
First, calculate the number of moles of benzene and toluene:
Number of moles of benzene = mass of benzene / molar mass of benzene
= 44.7 g / 78.11 g/mol
= 0.572 mol
Number of moles of toluene = mass of toluene / molar mass of toluene
= 77.3 g / 92.14 g/mol
= 0.839 mol
Next, calculate the mole fraction of benzene:
Mole fraction of benzene = moles of benzene / (moles of benzene + moles of toluene)
= 0.572 mol / (0.572 mol + 0.839 mol)
= 0.405
To calculate the partial vapor pressure of benzene, use Raoult's Law:
Partial vapor pressure of benzene = mole fraction of benzene * vapor pressure of benzene
= 0.405 * 95.1 mmHg
≈ 38.5 mmHg
b) To determine the partial vapor pressure of toluene, we can use the same approach as in part a.
Mole fraction of toluene = 1 - mole fraction of benzene (since we are calculating the partial pressure of toluene)
= 1 - 0.405
= 0.595
Partial vapor pressure of toluene = mole fraction of toluene * vapor pressure of toluene
= 0.595 * 28.4 mmHg
≈ 16.9 mmHg
c) The total vapor pressure of the solution is the sum of the partial vapor pressures of benzene and toluene:
Total vapor pressure of the solution = partial vapor pressure of benzene + partial vapor pressure of toluene
= 38.5 mmHg + 16.9 mmHg
≈ 55.4 mmHg
To find the answers to these questions, we need to first understand the concept of partial vapor pressure and how it relates to the amount of each component in a solution.
a) The partial vapor pressure of benzene above the solution can be obtained using Raoult's law. According to Raoult's law, the partial vapor pressure of a component in a solution is directly proportional to its mole fraction in the mixture.
To find the partial vapor pressure of benzene, we need to calculate the mole fraction of benzene in the mixture:
1. Determine the moles of benzene (C6H6) and toluene (C6H5CH3) using their respective molar masses. The molar mass of benzene is 78.11 g/mol, and the molar mass of toluene is 92.14 g/mol.
Moles of benzene = mass of benzene / molar mass of benzene
= 44.7 g / 78.11 g/mol
Moles of toluene = mass of toluene / molar mass of toluene
= 77.3 g / 92.14 g/mol
2. Calculate the total moles of the mixture:
Total moles = moles of benzene + moles of toluene
3. Calculate the mole fraction of benzene:
Mole fraction of benzene = moles of benzene / total moles
4. Use the mole fraction of benzene to find the partial vapor pressure of benzene using Raoult's law:
Partial vapor pressure of benzene = mole fraction of benzene * vapor pressure of benzene at 25°C
b) To find the partial vapor pressure of toluene, we can use the same approach as in part a.
1. Calculate the moles of benzene and toluene using their respective molar masses.
2. Calculate the total moles of the mixture.
3. Calculate the mole fraction of toluene.
4. Use the mole fraction of toluene to find the partial vapor pressure of toluene using Raoult's law:
Partial vapor pressure of toluene = mole fraction of toluene * vapor pressure of toluene at 25°C
c) The total vapor pressure of a solution is the sum of the partial vapor pressures of each component. Therefore, to find the total vapor pressure of the solution, we need to add the partial vapor pressures of benzene and toluene.
Total vapor pressure of the solution = partial vapor pressure of benzene + partial vapor pressure of toluene