Question

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 g reserpine is dissolved in g camphor, the freezing-point depression is ( for camphor is ). Calculate the molality of the solution and the molar mass of reserpine. Answer mol/kg;

Answer 1

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.