1) What is the vapor pressure at 25°C of a solution of 1.20g ofnephtalene C10H8 in 25.6g of benzone C6H6? The vapor pressure of pure benzene at 23°C is 86.0mmHg the vapor pressure of naphthalene can be neglected. Calculate the vapor pressure lowering of the solution.
2)A solution was prepared by dissolving 0.834g of sulfur, S8in 100.0g of acetic acid HC2H3O2. Calculate the freezing point and the boiling point of the solution.
1)
mols naphthalene = grams/molar mass
mols benzene = grams/molar
Xnaph = nnaphth/total mols.
delt P = Xnaphth*Po benzene.
self p= 0.11atm
1) To calculate the vapor pressure lowering of the solution, we can use Raoult's Law. According to Raoult's Law, the vapor pressure of a solvent over a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
First, we need to calculate the mole fraction of benzene in the solution:
Molar mass of benzene (C6H6) = (12.01 g/mol x 6) + (1.01 g/mol x 6) = 78.11 g/mol
Molar mass of naphthalene (C10H8) = (12.01 g/mol x 10) + (1.01 g/mol x 8) = 128.17 g/mol
moles of benzene = mass of benzene / molar mass of benzene = 25.6 g / 78.11 g/mol = 0.328 mol
moles of naphthalene = mass of naphthalene / molar mass of naphthalene = 1.20 g / 128.17 g/mol = 0.00936 mol
Mole fraction of benzene = moles of benzene / (moles of benzene + moles of naphthalene)
= 0.328 mol / (0.328 mol + 0.00936 mol) = 0.972
Now, we can calculate the vapor pressure lowering using Raoult's Law:
vapor pressure lowering = (1 - mole fraction of benzene) x vapor pressure of pure benzene
= (1 - 0.972) x 86.0 mmHg
= 0.028 x 86.0 mmHg
= 2.41 mmHg
Therefore, the vapor pressure lowering of the solution is 2.41 mmHg.
2) To calculate the freezing point and the boiling point of a solution, we can use the equations for freezing point depression and boiling point elevation, respectively.
Freezing point depression equation: ΔTf = -Kf * molality
Boiling point elevation equation: ΔTb = Kb * molality
First, let's calculate the molality (moles of solute per kilogram of solvent) of the sulfur solution:
Molar mass of S8 = (32.07 g/mol x 8) = 256.56 g/mol
moles of sulfur = mass of sulfur / molar mass of sulfur = 0.834 g / 256.56 g/mol = 0.00325 mol
mass of acetic acid = 100.0 g
molality = moles of sulfur / mass of acetic acid (in kg) = 0.00325 mol / 0.100 kg = 0.0325 mol/kg
Now, let's calculate the freezing point depression (ΔTf) using the given values of the cryoscopic constant (Kf) for acetic acid:
Kf = 3.9 °C/m (given)
ΔTf = -Kf * molality = -3.9 °C/m * 0.0325 mol/kg = -0.1267 °C
So, the freezing point of the solution is lowered by approximately 0.127 °C.
Next, let's calculate the boiling point elevation (ΔTb) using the given values of the ebullioscopic constant (Kb) for acetic acid:
Kb = 3.07 °C/m (given)
ΔTb = Kb * molality = 3.07 °C/m * 0.0325 mol/kg = 0.0997 °C
Hence, the boiling point of the solution is raised by approximately 0.100 °C.
To calculate the vapor pressure and vapor pressure lowering of the solution in question 1, 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 the component multiplied by its vapor pressure in the pure state.
First, let's calculate the number of moles of naphthalene (C10H8) and benzene (C6H6) in the solution.
Molar mass of naphthalene (C10H8) = 128.18 g/mol
Molar mass of benzene (C6H6) = 78.11 g/mol
Number of moles of naphthalene:
1.20g / 128.18 g/mol = 0.00936 mol
Number of moles of benzene:
25.6g / 78.11 g/mol = 0.328 mol
Now we can calculate the mole fraction of benzene:
Mole fraction of benzene:
0.328 mol / (0.00936 mol + 0.328 mol) = 0.972
Next, we need to calculate the vapor pressure of the solution using Raoult's Law. The vapor pressure of benzene in the pure state is given as 86.0 mmHg.
Vapor pressure of the solution:
0.972 * 86.0 mmHg = 83.71 mmHg
The vapor pressure lowering of the solution can be calculated by subtracting the vapor pressure of the solution from the vapor pressure of benzene in the pure state.
Vapor pressure lowering:
86.0 mmHg - 83.71 mmHg = 2.29 mmHg
Therefore, the vapor pressure at 25°C of the solution is 83.71 mmHg and the vapor pressure lowering is 2.29 mmHg.
For question 2, to calculate the freezing point and boiling point of the solution, we need to use the equations for freezing point depression and boiling point elevation.
Freezing point depression equation:
ΔTf = Kf * m
Boiling point elevation equation:
ΔTb = Kb * m
ΔTf represents the change in freezing point, ΔTb represents the change in boiling point, Kf is the cryoscopic constant, Kb is the ebullioscopic constant, and m is the molality of the solution.
First, let's calculate the moles of sulfur (S8) and acetic acid (HC2H3O2) in the solution.
Molar mass of sulfur (S8) = 256.52 g/mol
Molar mass of acetic acid (HC2H3O2) = 60.05 g/mol
Number of moles of sulfur:
0.834g / 256.52 g/mol = 0.00325 mol
Number of moles of acetic acid:
100.0g / 60.05 g/mol = 1.664 mol
Next, we can calculate the molality of the solution, which is the number of moles of solute per kilogram of solvent.
Mass of solvent (acetic acid):
100.0g = 0.100 kg
Molality (m) of the solution:
0.00325 mol / 0.100 kg = 0.0325 mol/kg
Now we need to find the cryoscopic constant (Kf) and ebullioscopic constant (Kb) for acetic acid. These constants represent the change in freezing point and boiling point per molal concentration of the solute.
Cryoscopic constant (Kf) for acetic acid = 3.90 °C/m
Ebullioscopic constant (Kb) for acetic acid = 3.07 °C/m
Using the freezing point depression equation and boiling point elevation equation, we can calculate the change in freezing point (ΔTf) and change in boiling point (ΔTb).
Change in freezing point (ΔTf):
ΔTf = 3.90 °C/m * 0.0325 mol/kg = 0.12675 °C
Change in boiling point (ΔTb):
ΔTb = 3.07 °C/m * 0.0325 mol/kg = 0.099775 °C
To calculate the actual freezing point and boiling point, we need to add the change in freezing point and boiling point to the normal freezing and boiling points of acetic acid.
Normal freezing point of acetic acid = 16.6 °C
Normal boiling point of acetic acid = 118.1 °C
Freezing point of the solution:
16.6 °C - 0.12675 °C = 16.47325 °C
Boiling point of the solution:
118.1 °C + 0.099775 °C = 118.199775 °C
Therefore, the freezing point of the solution is approximately 16.47325 °C and the boiling point of the solution is approximately 118.199775 °C.