Asked by ryancon
Tf (acetic acid) = 16.1°C
Tf (with 3.00g of solute in 10.00mL of acetic acid) = - 2.5°C
Use the data to calculate the molecular mass of the solute (Acetic acid: Kf = 3.9°C•kg/mol; D = 1.05 g/mL)
After finding molality I was completely stuck on what to do next..
Tf (with 3.00g of solute in 10.00mL of acetic acid) = - 2.5°C
Use the data to calculate the molecular mass of the solute (Acetic acid: Kf = 3.9°C•kg/mol; D = 1.05 g/mL)
After finding molality I was completely stuck on what to do next..
Answers
Answered by
DrBob222
OK, you have molality = m
m = mol/kg solvent. You have m and kg solvent, substitute and solve for mol.
Then mol = g/molar mass. You have mol and grams, solve for molar mass.
m = mol/kg solvent. You have m and kg solvent, substitute and solve for mol.
Then mol = g/molar mass. You have mol and grams, solve for molar mass.
Answered by
ryancon
Could you please explain the steps? I am still unsure about what I am doing...
Answered by
DrBob222
The steps are there. Just follow my instructions What's confusing?
.You say you have m. Then m = mols/kg solvent. The problem tells you kg solvent is 10 mL acetic acid. density of the acid iis 1.05 g/mL so
mass = density x volume = 1.05 g/mL x 10 mL = 10.5 grams or 0.0105 kg.
m = mols/0.0105 kg and mols = m x 0.0105 = ? mols. Plug in your m to find mols. Then mols = grams/molar mass. You have just calculated mols. The problem tells you grams = 3.00. Solve for molar mass.
.You say you have m. Then m = mols/kg solvent. The problem tells you kg solvent is 10 mL acetic acid. density of the acid iis 1.05 g/mL so
mass = density x volume = 1.05 g/mL x 10 mL = 10.5 grams or 0.0105 kg.
m = mols/0.0105 kg and mols = m x 0.0105 = ? mols. Plug in your m to find mols. Then mols = grams/molar mass. You have just calculated mols. The problem tells you grams = 3.00. Solve for molar mass.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.