Question

If \(52.5 \mathrm{g}\) of LiF is dissolved in \(306 \mathrm{g}\) of water, what is the expected freezing point of the solution? (Assume the van't Hoff factor, \(i\), for LiF is \(2 .\) )

Short answer

The expected freezing point of the solution is approximately \(-24.54\, °C\).

Step-by-step solution

Calculate the molality of the LiF solution

First, we need to find the molality (moles of solute per kilogram of solvent) of the LiF solution. - Find the molecular weight of LiF: - Lithium (Li): approximately 7 g/mol - Fluorine (F): approximately 19 g/mol - So, the molecular weight of LiF = 7 + 19 = 26 g/mol.- Calculate the number of moles of LiF: \[\text{Moles of LiF} = \frac{52.5 \text{ g}}{26 \text{ g/mol}} = 2.019 \text{ moles}\]- Convert the mass of water to kilograms:\[306 \text{ g water} = 0.306 \text{ kg water}\]- Calculate the molality: \[m = \frac{2.019 \text{ moles}}{0.306 \text{ kg}} = 6.599 \text{ mol/kg}\]

Determine the freezing point depression

We use the formula for freezing point depression \[\Delta T_f = i \cdot K_f \cdot m \]where:- \(i\) is the van't Hoff factor, for LiF, \(i = 2\)- \(K_f\) is the cryoscopic constant for water, approximately \(1.86\ °C \, kg/mol\)- \(m\) is the molality of the solution, calculated in Step 1.Substitute the values:\[\Delta T_f = 2 \times 1.86 \, °C/kmol \times 6.599 \, mol/kg = 24.542\, °C\]

Calculate the new freezing point of the solution

The normal freezing point of water is \(0\, °C\). To find the freezing point of the solution, subtract the freezing point depression from the normal freezing point of water:\[\text{Freezing Point of Solution} = 0\, °C - 24.542\, °C = -24.542\, °C\]

Key concepts

Molality

Molality is a way of expressing the concentration of a solution. It tells us how many moles of a solute are present per kilogram of the solvent. In our context, it's useful for understanding how a substance like lithium fluoride affects water when dissolved. Unlike molarity, molality is not affected by temperature changes because it depends solely on the mass of the solvent, not its volume.

Here's how you calculate it:

• First, determine the molecular weight of the solute. For LiF, the molecular weight is found by adding the atomic masses of lithium (approximately 7 g/mol) and fluorine (approximately 19 g/mol), giving a total of 26 g/mol.
• Next, calculate the number of moles of LiF by dividing the mass of the LiF by its molecular weight. In our case, 52.5 g ÷ 26 g/mol results in roughly 2.019 moles of LiF.
• Finally, convert the mass of the solvent (water) into kilograms for compatibility with molality's definition. With 306 g of water, this conversion means 0.306 kg.
• The resulting value of molality, using these values, becomes 6.599 mol/kg.

This consistent approach shows why molality is a preferred unit under conditions where temperature variation could lead to volume changes.

van't Hoff factor

The van't Hoff factor, often symbolized as "i," is crucial in colligative property calculations like freezing point depression. It's a measure of the effect of a solute on the properties of a solution due to the number of particles the solute separates into when dissolved.

For ionic compounds like LiF, the van't Hoff factor reflects the number of ions formed. LiF dissociates into lithium ions and fluoride ions in solution, giving an "i" value of 2. This means that each molecule of LiF results in two particles in the solution.

Understanding this concept is key because:

• The van't Hoff factor directly influences how much the boiling or freezing point of the solution changes.
• Greater "i" values lead to more pronounced changes in colligative properties. Hence the freezing point depression becomes larger with greater dissociation, as seen in the LiF-water solution example.

Always consider the van't Hoff factor when dealing with ionic solutes to predict the extent of colligative effects accurately.

Cryoscopic constant

The cryoscopic constant, represented by "K_f," is a property specific to each solvent. It signifies how much the freezing point of the solvent decreases per molal concentration of a non-volatile, non-electrolyte solute added. Essentially, "K_f" tells us how sensitive a solvent's freezing point is to added solutes.

In our example, water's "K_f" is approximately 1.86 °C kg/mol. This value is used in calculating freezing point depression using the formula:\[\Delta T_f = i \times K_f \times m\]where, \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor, and \(m\) is the molality.

Points to remember:

• Each solvent has a unique cryoscopic constant which must be used correctly in calculations.
• A higher cryoscopic constant means a larger impact on freezing point depression per mole of solute.
• In the solution discussed, the calculated freezing point depression of 24.542 °C using water's "K_f" shows the significant effect of the solute on lowering the freezing point.

Knowing the cryoscopic constant helps predict how a solution will behave under changes, making it vital in practical applications and theoretical work.