Question

An aqueous solution of LiX is prepared by dissolving \(3.58 \mathrm{~g}\) of the electrolyte in \(283 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}(d=1.00 \mathrm{~g} / \mathrm{mL})\) The solution freezes at \(-1.81^{\circ} \mathrm{C}\). What is \(\mathrm{X}^{-}\) ? (Assume complete dissociation of \(\mathrm{LiX}\) to \(\mathrm{Li}^{+}\) and \(\mathrm{X}^{-}\).)

Short answer

Answer: The anion X in the aqueous solution of LiX is the fluoride ion (F).

Step-by-step solution

Calculate the change in freezing point

First, we need to find the freezing point depression. The solution freezes at \(-1.81^{\circ}\mathrm{C}\), and the freezing point of pure water is \(0^{\circ}\mathrm{C}\) So, the change in freezing point, \(\Delta T_f\) is: \(\Delta T_f = T_f - T_{f\:pure\:water} = -1.81 - 0 = -1.81^{\circ}\mathrm{C}\)

Calculate the molality of the solution

Using the freezing point depression equation, we can find the molality (m) of the solution: \(\Delta T_f = K_f \times m \times i\) Where \(K_f\) is the freezing point depression constant for water (\(1.86 \frac{^{\circ}\mathrm{C} \cdot\mathrm{kg}}{\mathrm{mol}}\)), m is the molality of the solution, and i is the van't Hoff factor (the number of particles formed due to dissociation, which is 2 for LiX since it dissociates into \(\mathrm{Li^+}\) and \(\mathrm{X^-}\)). We can now solve for m: \(m = \frac{\Delta T_f}{K_f \times i} = \frac{-1.81}{1.86\times 2} = -0.486 \:\mathrm{mol}\:/\:\mathrm{kg\:}\mathrm{of\:}\mathrm{water}\)

Calculate the moles of LiX

To find the moles of LiX, we need to use the molality equation: \(m = \frac{moles\:of\:solute}{mass\:of\:solvent\:in\:kg}\) We can rearrange the equation to find the moles of LiX: \(moles\:of\:solute = m \times mass\:of\:solvent\:in\:kg = -0.486 \times 0.283 = -0.1375\: mol\)

Calculate the molar mass of LiX

To find the molar mass of LiX, use the equation: \(molar\:mass\:of\:LiX = \frac{mass\:of\:LiX}{moles\:of\:LiX} = \frac{-3.58}{-0.1375} = 26.0364 \:\mathrm{g/mol}\)

Determine the anion X

Subtract the molar mass of Li to find the molar mass of X: \(molar\:mass\:of\:X = molar\:mass\:of\:LiX - molar\:mass\:of\:Li = 26.0364 - 6.9394 = 19.097 \:\mathrm{g/mol}\) Looking at the periodic table, we can identify that the anion with a molar mass close to 19.097 g/mol is the fluoride ion (F), with a molar mass of 19.00 g/mol. Hence, X is the fluoride ion (F), and the compound LiX is lithium fluoride (LiF).

Key concepts

Molality Calculation

Understanding molality is essential when studying freezing point depression, a property of solutions. Molality, denoted as 'm', is a measure of the concentration of a solute in a solution. Unlike molarity, which depends on the volume of the solution, molality is based on the mass of the solvent, making it temperature-independent.

To calculate molality, you use the formula: \[ m = \frac{{\text{{moles of solute}}}}{{\text{{mass of solvent in kg}}}} \]. Given the mass of the solute and the solvent (in kilograms), you first determine the moles of the solute. This involves dividing the mass of the solute by its molar mass. Subsequently, you apply the molality formula to find the concentration in mol/kg.

For example, in a scenario where you have a certain mass of a substance dissolved in a known mass of water, you'd calculate the molality by first converting the mass of water to kilograms. You'd then divide the number of moles of the dissolved substance by this mass. This step is critical in colligative property calculations and helps us understand the solution's behavior under different conditions.

van't Hoff Factor

The van't Hoff factor, symbolized as 'i', plays a pivotal role in the study of colligative properties. It indicates the number of particles into which a compound dissociates in solution. For instance, sodium chloride (NaCl) dissociates into two ions, Na+ and Cl−, thus, its van't Hoff factor is 2.

In our exercise example, LiX dissociates into Li+ and X−, which also gives us a van't Hoff factor of 2. It's important to note that not all solutes dissociate completely, and the extent of dissociation can greatly influence the solution's colligative properties.

When you're working with the equation for freezing point depression: \[ \Delta T_f = K_f \times m \times i \], the van't Hoff factor 'i' is multiplied by the molality and the freezing point depression constant (Kf) to calculate the change in freezing point. It serves as a correction factor to account for the actual number of particles present in the solution.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the number of dissolved particles rather than their chemical identity. These include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Freezing point depression, the focus of our exercise, is one such property where the freezing point of a solution is lower than that of the pure solvent due to the presence of a solute. This is quantified by the equation: \[ \Delta T_f = K_f \times m \times i \], where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the solvent's freezing point depression constant, 'm' is the solution’s molality, and 'i' is the van’t Hoff factor.

In practical applications, measuring freezing point depression can help in determining molecular weights of solutes and the functioning of antifreeze in vehicles. For students, understanding how to manipulate the relevant formulas to solve problems is a valuable skill in chemistry and related subjects.