Problem 51 - Chemistry and Chemical Reactivity

Question

Phenylcarbinol is used in nasal sprays as a preservative. A solution of \(0.52 \mathrm{g}\) of the compound in \(25.0 \mathrm{g}\) of water has a melting point of \(-0.36^{\circ} \mathrm{C} .\) What is the molar mass of phenylcarbinol?

Step-by-Step Solution

Answer

The molar mass of phenylcarbinol is approximately 134.6 g/mol.

Step 1: Understanding Freezing Point Depression Formula

Freezing point depression is a colligative property given by the formula \[ \Delta T_f = i \cdot K_f \cdot m \]Where:- \(\Delta T_f\) is the freezing point depression, calculated as the difference between the freezing point of the pure solvent and the solution.- \(i\) is the van't Hoff factor for the solute (i.e., the number of particles the solute breaks into, which is 1 for non-electrolytes like phenylcarbinol).- \(K_f\) is the cryoscopic constant for the solvent, which is water (1.86 °C·kg/mol).- \(m\) is the molality of the solution.

Step 2: Calculate the Freezing Point Depression

Calculate the freezing point depression \(\Delta T_f\) by subtracting the freezing point of the solution \(-0.36^{\circ} \mathrm{C}\) from the freezing point of pure water, \(0.0^{\circ} \mathrm{C}\):\[\Delta T_f = 0.0^{\circ} \mathrm{C} - (-0.36^{\circ} \mathrm{C}) = 0.36^{\circ} \mathrm{C}\]

Step 3: Calculate Molality of the Solution

Rearrange the formula to solve for \(m\):\[m = \frac{\Delta T_f}{i \cdot K_f}\]Substitute the known values: \(\Delta T_f = 0.36^{\circ} \mathrm{C}\), \(K_f = 1.86 \mathrm{°C \cdot kg/mol}\), and \(i = 1\):\[m = \frac{0.36^{\circ} \mathrm{C}}{1 \cdot 1.86^{\circ} \mathrm {C \cdot kg/mol}} = 0.19355 \mathrm{mol/kg}\]

Step 4: Use Molality to Find Molar Mass

Molality \(m\) is defined as the moles of solute per kilogram of solvent. Rearrange the formula to solve for the molar mass \(M\):\[m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{\frac{\text{Mass of solute}}{M}}{\text{kg of solvent}}\]Rearrange to find the molar mass \(M\):\[M = \frac{\text{Mass of solute} \times \text{kg of solvent}}{m}\]Given the mass of the solute is \(0.52\, \mathrm{g} = 0.00052\, \mathrm{kg}\), the mass of the solvent is \(25.0 \mathrm{g} = 0.025 \mathrm{kg}\), and \(m = 0.19355 \mathrm{mol/kg}\):\[M = \frac{0.00052\, \mathrm{kg} \times 0.025\, \mathrm{kg}}{0.19355\, \mathrm{mol/kg}} = \frac{0.000013}{0.19355\, \mathrm{mol/kg}} = 134.6\, \mathrm{g/mol}\]

Step 5: Conclusion

The molar mass of phenylcarbinol, derived using the freezing point depression data, is approximately \(134.6\, \mathrm{g/mol}\).

Key Concepts

Colligative Properties

Colligative properties are fascinating because they depend on the number of solute particles in a solution rather than their particular identity. Freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure are some examples of colligative properties. Each of these properties alters according to the concentration of the solute particles. This means that even if the chemical nature of the solute changes, the effect on the solvent remains the same as long as the number of solute particles is unchanged.

For instance, when a non-electrolyte like phenylcarbinol is dissolved in water, molecules do not break apart. Therefore, the colligative effect like freezing point depression only depends on the total number of phenylcarbinol molecules in the solution, making calculations a bit more straightforward, as we do not have to account for ionization. This understanding helps us calculate freezing point depression effectively, showcasing an essential application of colligative properties.

Molar Mass Calculation

Calculating molar mass using freezing point depression is a practical laboratory skill that uses the change in freezing point of a solution to deduce the molecular weight of a solute. To execute this calculation, you need to know the depression in the freezing point, the cryoscopic constant (K_f), and the mass of both solute and solvent.

Start by calculating the freezing point depression, which is the difference between the freezing point of the pure solvent and the solution. Then use the formula for molality (mol/kg), inserting the depression data, and rearranging to solve for molar mass. This involves using the equation: \[ M = \frac{\text{Mass of solute} \times \text{kg of solvent}}{m} \] Here, "M" is the molar mass, "m" is molality, and you will calculate molality from the solution chemistry.

By applying this step-by-step, particularly with the solute phenylcarbinol and water as solvent, chemists can figure out the molar mass accurately. The process hinges on accurate measurements and a clear understanding of molal concentration and molar mass relationships.

Cryoscopic Constant

The cryoscopic constant, sometimes described as the molal freezing point depression constant ( K_f), is a specific property of each solvent that helps in determining how the freezing point of a solvent decreases when a non-volatile solute is added. It provides the change in freezing point per molal concentration of the solute added.

For instance, water, the solvent in our example with phenylcarbinol, has a cryoscopic constant of 1.86 °C·kg/mol. This means that for every mol of solute per kilogram of water, the freezing point drops by 1.86 °C.

This constant is crucial when solving problems involving freezing point depression, as it allows us to relate changes in physical properties directly to the number of solute particles. Understanding and utilizing the cryoscopic constant correctly ensures accurate calculations of both molality and consequently, the molecular mass of the solute.