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.
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!
See your post above.
Update for Answer: Molar Mass is 497 grams/mole
My apologies for the mistake. I misread my paper
To calculate the boiling point elevation (ΔTB) for a solution, we need to use the equation:
ΔTB = KB * m
Where:
- ΔTB is the boiling point elevation,
- KB is the molal boiling point elevation constant, and
- m is the molality of the solution.
To find the molality (m), we need to calculate the number of moles of the solute (the compound) and the mass of the solvent (carbon tetrachloride).
First, let's find the number of moles of the solute (the compound). We are given the mass of the compound (2.00 g) and its molar mass is not provided. To find the molar mass, we need the chemical formula of the compound. Without that information, we cannot proceed further with the calculation.