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 \(1.00 \mathrm{~g}\) reserpine is dissolved in \(25.0 \mathrm{~g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\) for camphor is \(40 .{ }^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) ). Calculate the molality of the solution and the molar mass of reserpine.

Short answer

The molality of the solution is 0.06575 mol/kg, and the molar mass of reserpine is approximately 608.24 g/mol.

Step-by-step solution

Calculate the molality of the solution using the freezing-point depression formula

The freezing-point depression formula is given by: ΔT = Kf × m × i Where: ΔT = freezing-point depression (given as 2.63°C) Kf = cryoscopic constant for the solvent (given as 40°C·kg/mol for camphor) m = molality of the solution i = van 't Hoff factor (assumed to be 1 for reserpine since it is a non-electrolyte) We can solve this equation for the molality (m) of the solution as follows: m = ΔT / (Kf × i) Now, let's calculate the molality by plugging in the given values: m = \( \frac{2.63}{40 \times 1} \) m = 0.06575 mol/kg

Determine the moles of reserpine

Now that we have the molality of the solution, we can determine the number of moles of reserpine by using the formula for molality: molality = moles of solute / mass of solvent (in kg) Let's denote the number of moles of reserpine as n. Then we can write the following equation: 0.06575 mol/kg = n / (0.025 kg) Now, let's solve for n: n = 0.06575 × 0.025 n = 0.00164375 moles

Calculate the molar mass of reserpine

To calculate the molar mass of reserpine, we can use the formula: molar mass = mass of solute / moles of solute The mass of reserpine is given as 1.00 g. Therefore, we can calculate the molar mass as follows: molar mass = \( \frac{1.00}{0.00164375} \) molar mass = 608.24 g/mol Thus, the molar mass of reserpine is approximately 608.24 g/mol.

Key concepts

Molality Calculation

Molality is a measure of the concentration of a solute in a solution, which is particularly important in freezing-point depression studies. Unlike molarity, which depends on the volume of the solution, molality is based on the mass of the solvent and remains unaffected by temperature changes.

To calculate molality, we use the formula: \( molality = \frac{moles\ of\ solute}{mass\ of\ solvent\ (kg)} \).

It's essential to correctly convert the mass of the solvent from grams to kilograms when performing this calculation, as molality is expressed in terms of moles per kilogram of solvent. To find the 'moles of solute', we need to know or calculate the molar mass of the solute, which brings us to the importance of accurately determining a compound's molar mass for the molality calculation.

Cryoscopic Constant

The cryoscopic constant (\( K_f \)), is a unique physical property of the solvent, which is crucial in the formula that connects molality to freezing-point depression. It is defined as the depression of the freezing point of a solvent when one mole of a non-volatile solute is dissolved in one kilogram of the solvent.

Importance in Freezing-Point Depression

Freezing-point depression is a colligative property, meaning it depends on the number of particles of solute in a solution, rather than the type of particles. The cryoscopic constant represents the sensitivity of the solvent's freezing point to the presence of solute particles. For example, camphor has a cryoscopic constant of \( 40^{circ}C\cdot kg/mol \), which is applied in the calculation to find the molality.

Understanding the cryoscopic constant is essential for solving problems involving freezing-point depression, and it emphasizes why different solvents will have different effects on the freezing point, even with the same solute.

Molar Mass Determination

Determining molar mass is a fundamental concept in chemistry and is essential in calculating molality for freezing-point depression experiments. The molar mass is the weight of one mole (Avogadro's number of particles) of a substance and is typically expressed in grams per mole (\( g/mol \)).

To find the molar mass from an experimental freezing-point depression, we rearrange the formula \( molar\ mass = \frac{mass\ of\ solute}{moles\ of\ solute} \), which allows us to calculate the molar mass of the solute using the mass of the solute and the number of moles we derived from finding the molality.

When dealing with freezing-point depression, the molar mass can be back-calculated if we have the mass of the solute and the molality of the solution, as highlighted in our example with reserpine. Accurate molar mass values are crucial for chemists to understand the substance's stoichiometry and to predict the outcomes of chemical reactions and physical processes like freezing-point depression.