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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.

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.

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.