Question

What mass of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right),\) a nonelectrolyte, must be dissolved in \(200.0 \mathrm{g}\) water to give a solution with a freezing point of \(-1.50^{\circ} \mathrm{C} ?\)

Short answer

To obtain a solution with a freezing point of -1.50˚C, approximately 14.84 g of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) must be dissolved in 200.0 g of water.

Step-by-step solution

Calculate the change in freezing point (∆Tf)

The change in freezing point is the difference between the freezing point of the solution and the freezing point of the pure solvent. For this problem, the freezing point of the solution is given as -1.50˚C and the freezing point of pure water is 0˚C. ∆Tf = (Freezing point of pure solvent) - (Freezing point of the solution) ∆Tf = 0 - (-1.50˚C) ∆Tf = 1.50˚C

Calculate the molality of the glycerin solution

Now that we have the change in freezing point and the freezing point constant for water, we can use the freezing point depression formula to find the molality of the glycerin solution. The formula is: ∆Tf = Kf * molality Here, ∆Tf = 1.50˚C and Kf = 1.86 ˚C/m 1.50˚C = 1.86˚C/m * molality Molality = 1.50˚C / 1.86˚C/m Molality ≈ 0.806 moles glycerin/kg water

Calculate the moles of glycerin needed

Now we know the molality of the glycerin solution, we can compute the number of moles of glycerin needed to be dissolved in 200 g water. moles of glycerin = molality * mass of water (in kg) moles of glycerin = 0.806 mol/kg * 0.200 kg moles of glycerin ≈ 0.1612 mol

Calculate the mass of glycerin needed

Now we know the moles of glycerin, we can calculate their mass using the formula: mass = moles * molar mass The molar mass of glycerin (C₃H₈O₃) is Molar mass = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) + (3 * 16.00 g/mol) ≈ 92.09 g/mol mass of glycerin = 0.1612 mol * 92.09 g/mol mass of glycerin ≈ 14.84 g So, to get a solution with a freezing point of -1.50˚C, approximately 14.84 g of glycerin must be dissolved in 200.0 g of water.

Key concepts

Colligative Properties

Colligative properties are characteristics of solutions that depend solely on the number of dissolved particles in the solution, not on the identity of those particles. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. Freezing point depression, the focus of our textbook exercise, is particularly important in many practical applications, such as anti-freeze solutions for cars and the salting of roads in winter.

When a non-volatile solute like glycerin is dissolved in a solvent like water, it disrupts the formation of ice at 0°C, causing the solution to freeze at a lower temperature. This is because the solute particles interfere with the ability of the solvent molecules to form a crystalline solid (ice) by getting in between them, requiring a lower temperature to achieve the same level of solid formation.

Molality

Molality is a measure of the concentration of a solution that expresses the moles of solute per kilogram of solvent. It is distinct from molarity, which is the moles of solute per liter of solution. Molality is particularly useful in the study of colligative properties because it is not affected by temperature changes, unlike molarity. This is because the molality is based on mass, a property that doesn't change with temperature, while volume, used in molarity, can expand or contract with temperature.

To calculate molality, you divide the number of moles of solute by the mass of the solvent in kilograms. In our exercise, we calculated the molality to determine the amount of glycerin needed to lower the freezing point of water. This concept is crucial for understanding how solutes affect the physical properties of solutions in a temperature-dependent manner.

Solution Concentration

Solution concentration is a quantitative measure of how much solute is present in a given quantity of solvent or solution. It can be expressed in various ways, including molality, molarity, parts per million, and weight percent. Understanding concentration is fundamental in chemistry because it affects reaction rates, chemical equilibria, and physical properties of solutions.

In the context of our textbook problem, we focused on the concentration expressed as molality to utilize the linear relationship between the freezing point depression and the molality of the solution. Knowing the concentration allows us to predict and control the freezing point of the solution to suit particular needs, such as creating a solution with a specific freezing point.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a single molecule of the substance. The molar mass is an important factor when converting between the mass of a substance and the number of moles of that substance.

In our solution, we used the molar mass of glycerin to find the mass required to achieve the desired freezing point depression. The molar mass enabled us to convert moles (calculated from molality and mass of the solvent) into grams of glycerin needed. A correct molar mass is essential for accurately determining the amount of solute in a solution and for ensuring desired outcomes in chemical reactions and processes.