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 \(2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\) 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 reserpine and camphor solution is approximately \(0.06575\, \mathrm{mol/kg}\), and the molar mass of reserpine is approximately \(608.69\,\mathrm{g/mol}\).

Step-by-step solution

Write down the freezing-point depression formula

The freezing-point depression formula is given by: \[\Delta T_{f} = K_{f} \cdot m \cdot i\] where \(\Delta T_{f}\) is the freezing-point depression, \(K_{f}\) is the molal freezing-point depression constant (also known as the cryoscopic constant), \(m\) is the molality of the solution, and \(i\) is the van't Hoff factor (number of particles produced by a formula unit). In this case, since both reserpine and camphor are non-electrolytes, the van't Hoff factor is 1.

Calculate the molality of the solution

We are given that \(\Delta T_{f} = 2.63^{\circ}\mathrm{C}\), and the cryoscopic constant for camphor \(K_{f} = 40.0^{\circ}\mathrm{C}\cdot\mathrm{kg}/\mathrm{mol}\). Solving for the molality, we have: \[m = \frac{\Delta T_{f}}{K_{f} \cdot i} = \frac{2.63}{40.0 \cdot 1}\] Calculate the molality: \[m \approx 0.06575\, \mathrm{mol/kg}\]

Calculate the molar mass of reserpine

Now that we have the molality of the solution, we can use it to find the molar mass of reserpine. We have 1.00 g of reserpine dissolved in 25.0 g of camphor. First, convert the mass of camphor to kg: \[25.0\,\mathrm{g} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 0.025\,\mathrm{kg}\] Now, using the molality, we can find the number of moles of reserpine in the solution: \[0.06575\,\mathrm{mol/kg} \times 0.025\,\mathrm{kg} \approx 0.00164375\,\mathrm{mol}\] Finally, we can find the molar mass of reserpine by dividing its mass by the number of moles: \[\frac{1.00\,\mathrm{g}}{0.00164375\,\mathrm{mol}} \approx 608.69\,\mathrm{g/mol}\] So, the molar mass of reserpine is approximately 608.69 g/mol.

Key concepts

Molality Calculation

When discussing freezing-point depression, molality is a key concept. It measures the concentration of a solute in a solution. Unlike molarity, which is affected by changes in volume due to temperature changes, molality remains constant because it depends on the mass of the solvent rather than the volume. Molality is calculated using the formula:

• \[ m = \frac{n_{ ext{solute}}}{m_{ ext{solvent}}} \]

where \(m\) is the molality, \(n\) is the number of moles of solute, and \(m_{\text{solvent}}\) is the mass of the solvent in kilograms.
This makes it particularly useful in colligative properties like freezing-point depression, where temperature influences don't need to distort concentration measurements.
To calculate molality in the exercise, the freezing-point depression formula was used: \[ \Delta T_{f} = K_{f} \times m \times i \]. Given \( \Delta T_{f} = 2.63^{\circ}\mathrm{C} \) and \( K_{f} = 40.0^{\circ}\mathrm{C}\cdot\mathrm{kg}/\mathrm{mol} \), the van’t Hoff factor \(i\) is 1 since both reserpine and camphor do not dissociate into ions. Therefore:

• \[ m = \frac{2.63}{40.0} = 0.06575\, \mathrm{mol/kg} \]

This calculation gives us the molality of the solution.

Molar Mass Determination

Once the molality of a solution is calculated, it can be used to determine the molar mass of a solute, such as reserpine, within the solution. Here's how that process unfolds step-by-step:

• First, understand the relationship: the molar mass of a substance \(M\), is the mass of that substance divided by the number of moles it contains. The formula is:
  \[ M = \frac{m_{\text{solute}}}{n_{\text{solute}}} \]
• Here, \(m_{\text{solute}}\) is the mass of the solute, and \(n_{\text{solute}}\) is obtained from the molality multiplication:
• Convert camphor's mass to kilograms: \[ 25.0\,\mathrm{g} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 0.025\,\mathrm{kg} \]
• Calculate moles of reserpine using molality: \[ 0.06575\,\mathrm{mol/kg} \times 0.025\,\mathrm{kg} \approx 0.00164375\,\mathrm{mol} \]
• Finally, determinethe molar mass:\[ M \approx \frac{1.00\,\mathrm{g}}{0.00164375\,\mathrm{mol}} = 608.69\,\mathrm{g/mol} \]

In this way, you find the molar mass of reserpine to be approximately 608.69 g/mol, which is a crucial step in identifying or verifying substances in chemistry.

Cryoscopic Constant

The cryoscopic constant, often symbolized as \( K_{f} \), is vital for calculating the freezing-point depression of solutions. It represents how much the freezing point of a solvent decreases when one mole of solute is dissolved in one kilogram of solvent.

• This constant is specific to each solvent and helps in understanding how that particular solvent behaves when substances are dissolved in it.
  
• For example, in the context of camphor, \( K_{f} \), the cryoscopic constant is given as \( 40.0^{\circ}\mathrm{C}\cdot\mathrm{kg}/\mathrm{mol} \). This value tells you how sensitive camphor is to the addition of solute in terms of its freezing-point lowering.
  
• In our calculation, this constant allowed us to use the formula \[ \Delta T_{f} = K_{f} \times m \times i \] effectively.
  

Understanding the cryoscopic constant helps to not just predict freezing-point depressions, but also relate them to the composition and properties of the solutions involved. Knowing the cryoscopic constant is key to solving many real-world problems in chemistry, especially those related to colligative properties.