What would be the freezing and boiling points of a solution prepared by
dissolving 1.00g of benzoic acid (C6H5CO2H) in 10.0g of benzene?
mols acid = grams/molar mass
m acid = mols/kg solvent
delta T = Kf*m
Solve for T and add to normal freezing point of benzene to find the new f.p.
Do the same for b.p.
mols acid is same
m acid is same
delta T = Kb*m
Add delta T to normal b.p. to find new b.p.
To determine the freezing and boiling points of the solution prepared by dissolving 1.00g of benzoic acid in 10.0g of benzene, you need to use the concept of colligative properties, specifically freezing point depression and boiling point elevation.
First, you should find the molality (m) of the solution, which is the amount of solute (in moles) divided by the mass of the solvent (in kilograms).
To calculate the molality, you need to convert the mass of the solute (benzoic acid) to moles and the mass of the solvent (benzene) to kilograms.
1. Calculate the moles of benzoic acid:
- The molar mass of benzoic acid (C6H5CO2H) is:
6(12.01 g/mol) + 5(1.01 g/mol) + 2(16.00 g/mol) + 1(12.01 g/mol) = 122.13 g/mol
- Convert the mass of benzoic acid to moles:
1.00 g / 122.13 g/mol = 0.0082 mol
2. Convert the mass of benzene to kilograms:
- 10.0 g = 10.0 / 1000 = 0.01 kg
3. Calculate the molality (m) of the solution:
- m = moles of solute / mass of solvent (in kg)
m = 0.0082 mol / 0.01 kg = 0.82 mol/kg
Once you have the molality, you can use the following formulas:
1. Freezing point depression (ΔTf):
- ΔTf = Kf * m
In this case, you need to know the freezing point depression constant (Kf) of benzene. Look it up in a reference book or online. For benzene, Kf is approximately 5.12 °C/molal.
2. Boiling point elevation (ΔTb):
- ΔTb = Kb * m
In this case, you need to know the boiling point elevation constant (Kb) of benzene. Look it up in a reference book or online. For benzene, Kb is approximately 2.53 °C/molal.
By plugging in the values for Kf, Kb, and m in the above equations, you can find the freezing point depression and boiling point elevation of the solution. Subtract or add the obtained values from the normal freezing and boiling points of benzene to get the freezing and boiling points of the solution, respectively.