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 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63°C ( K_f for camphor is$\mathbf{40}\mathbf{°}\mathbf{C}$kg/mol). Calculate the molality of the solution and the molar mass of reserpine.

Short answer

The depression in freezing point is given by the relation: $\Delta {\text{F=K}}_{\text{f}}\times \text{m}$, by which the molality of the solution will come 0.06575 and molar mass of solute that is reserpine will be 608.3

Step-by-step solution

Step 1:

It is given that

mass of reserpine is 1.00 g

mass of camphor is 25.0 g

Freezing point depression =2.63°C

And the value of K_f for camphor is $\text{40}°\text{C}$ kg/mol

Step 2:

The freezing point depression and molality of a compound are related to each as:

$\Delta {\text{F=K}}_{\text{f}}\times \text{m}$ or it can be written as $\text{m=}\frac{\Delta \text{F}}{{\text{K}}_{\text{f}}}$,

Thus, molality will be $\text{m=}\frac{2.63}{40}\text{=0.06575}$.

Step 3:

Molality of compound is given by the ratio of number of moles of solute to the volume of solvent in kilograms.

$\text{Molality=}\frac{\text{Massofsolute}\times \text{1000}}{\text{Molecularmassofsolute}\times \text{Volumeofsolvent}}$

So, by putting value we will get

$\begin{array}{l}\text{0.06575=}\frac{\text{1}\times \text{1000}}{\text{Molecularmassofsolute}\times \text{25}}\\ \text{Molecularmass=}\frac{1000}{\text{25}\times \text{0.06575}}\text{=608.3g}\end{array}$