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 \(25.0 \mathrm{g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}(K_{\mathrm{f}}\) for camphor is \(40 .^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol})\) Calculate the molality of the solution and the molar mass of reserpine.

Molality is a measure of concentration that expresses the moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality is only affected by the mass of the solvent, making it a valuable concentration measure, especially when dealing with temperature changes.

One molal solution contains one mole of the solute dissolved in one kilogram of solvent. In the context of our exercise, we calculated the molality of a reserpine solution by using the formula:
\[m = \frac{{\text{{moles of solute}}}}{{\text{{kilograms of solvent}}}}\].
Recognizing that the solute in our scenario is reserpine and the solvent is camphor, we can understand the steps needed to compute the molality. After finding the number of moles of reserpine and dividing it by the mass of the camphor in kilograms, we arrived at the solution's molality.

Colligative properties are characteristics of solutions that depend on the number of dissolved particles in a solvent and not on the nature of those particles. Freezing-point depression is a type of colligative property where the freezing point of the solvent decreases when a solute is dissolved in it.

This phenomenon occurs because the solute particles interfere with the formation of the solvent's crystalline structure, thus requiring a lower temperature to reach the solid state. The freezing-point depression formula that relates this concept with molality is:
\[\Delta T_f = K_f \cdot m\],
where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the cryoscopic constant (unique to each solvent), and \(m\) is the molality of the solution. In our reserpine and camphor solution, by knowing \(\Delta T_f\) and \(K_f\), we were able to calculate the molality. This example beautifully illustrates how a colligative property can be used to determine a solution's concentration.

The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (\(g/mol\)). It is a fundamental property used in chemistry to convert between the mass of a sample and the number of moles it contains, which is essential for conducting stoichiometric calculations.

In the given exercise, we utilized the molar mass to find the molecular weight of reserpine. After calculating the number of moles, the molar mass was determined by the following relationship:
\[\text{{Molar mass}} = \frac{{\text{{mass of solute (g)}}}}{{\text{{moles of solute}}}}\].
This equation shows that if we know the mass of the solute and the number of moles, we can find its molar mass. From the solution steps, we learned that despite the small quantity of reserpine used, its high molar mass translates to a significant effect on the freezing point of the solution.