I'm really confused on this question so if someone could show the steps and how to get the answer, that would be great. I understand that you would use delta T=Kbm, but don't know how to use the equation to find the answer. Thanks!!
Calculate the boiling point of a solution containing 7.58 grams of C10H8 in 120.0 grams of benzene. The boiling point of benzene is 80.1 degrees Celcius and the Kb = 2.52 degrees celcius/molal.
Note the correct spelling of celsius.
mols C10H8 = grams/molar mass = 7.58/molar mass C10H8 = ?
Then molality = m = mols/kg solvent
m = mols from above/0.120 = ?
Then delta T = Kb*m. You know Kb and m, solve for delta T, then add that to the normal boiling point of benzene in the problem to find the new boiling point.
I know what I did wrong. Some reason I divided the moles of C10H8 by .0120 instead of .120, so my answer was 11 times the answer it should have been. Thank you very much!
To find the boiling point of the solution, you can use the equation ΔT = K_b * m. Let's break down the steps to get the answer:
1. First, calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is C10H8 and the solvent is benzene.
To find the number of moles of C10H8, you divide the given mass of C10H8 by its molar mass.
Molar mass of C10H8 = (10 * atomic mass of C) + (8 * atomic mass of H)
= (10 * 12.01 g/mol) + (8 * 1.01 g/mol)
= 120.10 g/mol + 8.08 g/mol
= 128.18 g/mol
Number of moles of C10H8 = mass of C10H8 / molar mass
= 7.58 g / 128.18 g/mol
≈ 0.059 mol
Next, find the mass of benzene in kilograms:
Mass of benzene = 120.0 g / 1000
= 0.120 kg
Now we can calculate the molality:
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.059 mol / 0.120 kg
= 0.492 mol/kg
2. Now that we have the molality (m), we can use it in the equation ΔT = K_b * m to find the change in boiling point (ΔT). The boiling point of pure benzene is given as 80.1 degrees Celsius, and the value of K_b (the molal boiling point elevation constant) is given as 2.52 degrees Celsius/molal.
ΔT = K_b * m
= 2.52 degrees Celsius/molal * 0.492 mol/kg
≈ 1.24 degrees Celsius
3. Finally, we can find the boiling point of the solution by adding the change in boiling point to the boiling point of pure benzene:
Boiling point of solution = Boiling point of pure benzene + ΔT
= 80.1 degrees Celsius + 1.24 degrees Celsius
≈ 81.34 degrees Celsius
Hence, the boiling point of the solution containing 7.58 grams of C10H8 in 120.0 grams of benzene is approximately 81.34 degrees Celsius.