A 2.00 g sample of a particular compound was dissolved in 15.0 g of carbon tetrachloride. The boiling point of this solution was determined to be 77.85ºC.
For pure CCl4, TB = 76.50ºC and KB = 5.03°C·kg/mol
Calculate the boiling point elevation, ΔTB, for this solution.
3 answers
See your post above.
Answer: 498 grams/mole
(77.85-76.50)/5.03 = 0.268 kg/mole
(15.0 g(1 kg/1000 g) = 0.0150 kg
(0.268 mole/kg)(0.0150 kg) = 0.00402 moles
Molecular Weight = 2.00 g/0.00402 moles = 498 g/mole
I hope this quick step-by-step help as much as you or anyone would hope for! GL!
(77.85-76.50)/5.03 = 0.268 kg/mole
(15.0 g(1 kg/1000 g) = 0.0150 kg
(0.268 mole/kg)(0.0150 kg) = 0.00402 moles
Molecular Weight = 2.00 g/0.00402 moles = 498 g/mole
I hope this quick step-by-step help as much as you or anyone would hope for! GL!
Update for Answer: Molar Mass is 497 grams/mole
My apologies for the mistake. I misread my paper
My apologies for the mistake. I misread my paper
6 answers