Question

The freezing point (in \(\mathrm{C}\) ) of a solution containing \(0.1 \mathrm{~g}\) of \(\left.\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\right]\) (Mol. Wt. 329 ) in \(100 \mathrm{~g}\) of water \(\left(\mathrm{K}_{1}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\) is (A) \(-2.3 \times 10^{-2}\) (B) \(-5.7 \times 10^{-2}\) (C) \(-5.7 \times 10^{-3}\) (D) \(-1.2 \times 10^{-2}\)

Short answer

The freezing point of the solution is \( -5.7 \times 10^{-3} \degree C \) (Option C).

Step-by-step solution

Calculate the number of moles of solute

To calculate the moles of \(K_3Fe(CN)_6\) dissolved, use the formula moles = mass / molar mass. The mass of solute is 0.1g and the molar mass is 329g/mol.

Use the freezing point depression equation

The freezing point depression \(\Delta T_f\) is found using the equation \(\Delta T_f = K_f \times m\), where \(K_f\) is the cryoscopic constant and m is the molality of the solution.

Calculate molality

Molality (m) is the number of moles of solute per kilogram of solvent: \(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\).

Determine the freezing point depression

Substitute the values of molality and \(K_f\) into the freezing point depression equation to calculate \(\Delta T_f\).

Calculate the new freezing point

Subtract the freezing point depression \(\Delta T_f\) from the normal freezing point of water (0\degree C) to get the freezing point of the solution.

Match the answer with the provided options

Match the calculated freezing point to one of the given options (A), (B), (C), or (D).

Key concepts

Cryoscopic Constant

Understanding the cryoscopic constant is crucial when studying the freezing point depression of solutions. The cryoscopic constant, often represented by the symbol \(K_f\), is actually a proportionality constant that relates the freezing point depression of a solvent to the molality of a solute that is dissolved in it. The concept becomes clearer through the formula \(\Delta T_f = K_f \times m\), where \(\Delta T_f\) represents the depression in freezing point, and \(m\) stands for the molality of the solution.

The constant's value depends on the properties of the solvent, not the solute. For water, which is a common solvent in such calculations, \(K_f\) typically has a value of 1.86 Kelvin-kilograms per mole. Therefore, in calculating the freezing point depression, knowing the precise value of the cryoscopic constant for your specific solvent is a necessity to ensure accurate results.

In practical terms, the cryoscopic constant enables us to predict how much the freezing point of a liquid will lower when a solute is dissolved in it, which is absolutely vital in formulating antifreeze solutions and understanding natural phenomena like why salt is spread on icy roads.

Molality

Molality is another essential concept in the realm of solution chemistry, especially when it comes to colligative properties such as freezing point depression. It measures the concentration of a solution, defined as the number of moles of solute per kilogram of solvent. It is important to emphasize that molality is concerned with the mass of the solvent, not the volume of the solution. This distinction is particularly relevant because the volume can change with temperature, whereas the mass remains constant.

To calculate molality, you can use the formula \(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\). It's also beneficial to know that molality is used over molarity in freezing point depression calculations because it provides a more accurate representation of concentration in systems where temperature changes are involved, as it is independent of temperature.

Colligative Properties

Colligative properties are those properties of solutions that depend upon the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. Freezing point depression is one of these properties, along with boiling point elevation, vapor pressure lowering, and osmotic pressure. These properties are \(colligative\) because they collect together and depend solely on the number of solute particles irrespective of their type.

When a non-volatile solute is added to a pure solvent, the freezing point of the solution is lower than that of the pure solvent. This is because the addition of solute particles disrupts the formation of the solid structure of the solvent, effectively requiring a lower temperature to achieve the same solid structure.

It's pivotal for students to grasp that it is the number of particles that matter, which is why ionizing solutes can have a disproportionally large effect compared to non-ionizing ones. An example is \(K_3Fe(CN)_6\), a compound that dissociates into four particles upon dissolution; therefore, affecting the freezing point of water significantly, although only a small amount is dissolved.