Respuesta :
Answer:
- Molality = 0.066 m
- Molar mass = 608.36 g/mol
Explanation:
It seems the question is incomplete. However a web search us shows this data:
" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "
The freezing-point depression is expressed by:
- ΔT=Kf * m
We put the data given by the problem and solve for m:
- 2.63 °C = 40°C·kg/mol * m
- m = 0.06575 m
For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:
- 0.06575 m = Moles reserpine / kg camphor
- 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor
We calculate moles of reserpine:
- 0.06575 m = Moles reserpine / 0.025 kg camphor
- Moles reserpine = 1.64x10⁻³ mol
Finally we use the mass of reserpine and the moles to calculate the molar mass:
- 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol
Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.