Problem 131 - Chemistry An Atoms First Approach

Question

An aqueous solution containing 0.250 mole of Q, a strong electrolyte, in \(5.00 \times 10^{2} \mathrm{g}\) water freezes at \(-2.79^{\circ} \mathrm{C}\). What is the van't Hoff factor for Q? The molal freezing-point depression constant for water is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} .\) What is the formula of \(\mathrm{Q}\) if it is \(38.68 \%\) chlorine by mass and there are twice as many anions as cations in one formula unit of Q?

Step-by-Step Solution

Answer

The van't Hoff factor for Q is 3. The molar mass of Q is 70.90 g/mol. Thus, the formula of Q is H2Cl2 or HCl·HCl.

Step 1: Calculate the molality of the solution.

To calculate the molality of the solution, we'll use the formula \[ molality = \frac{moles \:of\: solute}{mass \:of\: solvent(kg)} \] We are given that the solution contains 0.250 mole of Q in 5.00*10^2 g of water (0.500 kg of water). Plugging these values into the formula, we get: \[ molality = \frac{0.250\: mole}{0.500\: kg} =0.500\: mol\cdot kg^{-1} \]

Step 2: Determine the freezing-point depression and find the van't Hoff factor.

We're given that the freezing-point depression constant for water (Kf) is 1.86 C kg/mol and that the solution freezes at -2.79 C. We'll use these values and the molality we calculated earlier to solve for the van't Hoff factor (i) using the freezing-point depression equation: \[ \Delta T_f = i \cdot K_f \cdot molality \] Rearranging the formula for i and plugging in the values, we get: \[ i = \frac{\Delta T_f}{K_f \cdot molality} = \frac{-(-2.79\: ^\circ{C})}{(1.86\: ^\circ{C}\cdot kg \cdot mol^{-1})(0.500 \: mol\cdot kg^{-1})} = 3 \] The van't Hoff factor for Q is 3.

Step 3: Calculate the molar mass of Q.

Using the van't Hoff factor (i), we can determine the molar mass of Q. Since i=3, the strong electrolyte Q dissociates into 3 ions in solution, so: \[ i \cdot moles \:of\: Q = total\: moles\: of\: ions \] We know that there are 0.250 moles of Q in the solution; plugging in the values, we get: \[ 3 \cdot moles\: of\: Q = 0.750\: moles\: of\: ions \] Now, we can find the molar mass of Q using the fact that we have twice as many anions as cations, so for every 1/3 moles of cations, we have 2/3 moles of anions. \[ Molar\: mass\: of\: Q = \frac{1}{0.75} \times (mass\: of\: cations + mass \: of\: anions) \] Since Q has 38.68 % chlorine by mass and there are twice as many anions as cations, we can rewrite the molar mass of Q formula in terms of chlorine mass as : \[ Molar\: mass\: of\: Q = \frac{1}{0.75} \times (\frac{1}{3} \times mass\: of\: Cl + 2 \times \frac{1}{3}\times mass \: of\: Cl) \] Plugging in 35.45 g/mol for the mass of Cl, we get: \[ Molar\: mass\: of\: Q = \frac{1}{0.75} \times (\frac{1}{3} \times 35.45 + 2 \times \frac{1}{3}\times 35.45) = 70.90\: g/mol \]

Step 4: Determine the formula of Q.

Using the molar mass and the mass percentage of chlorine, we can now find the formula of Q. To do this, let's use the following variables: 1 mole of cations = \(x\) g/mol 1 mole of anions = 35.45 g/mol (mass of Cl) Since there are twice as many anions as cations in one formula unit of Q, Molar mass of Q = \(x\) + 2(35.45) We previously calculated the molar mass of Q to be 70.90 g/mol, so 70.90 = \(x\) + 2(35.45) Solving for \(x\), we get: \(x\) = 70.9 - 70.9 = 0 g/mol This indicates that the cations are hydrogen ions. Since there are twice as many anions as cations in one formula unit of Q, and the cations are hydrogen ions (H+): The formula of Q is H2Cl2 or HCl·HCl.

Key Concepts

Molal Freezing-Point Depression

Understanding the concept of molal freezing-point depression is crucial for solving problems related to the freezing temperatures of solutions. In essence, the freezing point of a solution is lower than that of the pure solvent due to the addition of a solute. This phenomenon is quantified by the freezing-point depression equation:
ewlineewline \[ewline ewline \triangle T_f = i \times K_f \times mewline ewline \]where ewline ewline \(\triangle T_f\) is the change in freezing temperature, ewline ewline \(i\) is the van't Hoff factor that represents the number of particles the solute breaks into, ewline ewline \(K_f\) is the molal freezing-point depression constant specific to the solvent, and ewline ewline \(m\) stands for the molality, or moles of solute per kilogram of solvent.ewline ewline To find the change in freezing temperature of the solution, you multiply these three variables together. This concept is widely used in chemistry to determine the extent of ionization or dissociation of solutes in solution. In educational exercises, students are often asked to calculate the freezing-point depression to deduce the van't Hoff factor, which can then help identify the solute's properties or its formula, as seen in the exercise given.

Strong Electrolyte

A strong electrolyte is a substance that completely dissociates into ions when dissolved in water. Typical examples include salts like sodium chloride (NaCl) and acids like hydrochloric acid (HCl).
ewline ewline Understanding strong electrolytes is invaluable when analyzing the van't Hoff factor, as it directly influences the factor's value. The complete dissociation of an electrolyte means the number of particles in solution is significant, which in turn, greatly affects boiling point elevation and freezing point depression. This concept was applied in the step by step solution of the given exercise, where the solute Q was assumed to fully dissociate, which justified the calculation of a van't Hoff factor greater than 1.
ewline ewline In instructional content, it's important to clarify that the extent of dissociation is a key factor when using colligative properties to deduce information about the solute, something that challenges many students. When a solute is a strong electrolyte, the impact on the solution's physical properties should be proportional to the number of ions formed, leading to notable changes in properties such as freezing-point depression.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of that substance and is typically expressed in grams per mole (g/mol). Calculating the molar mass is a vital step when determining the molecular formula of a compound, especially when dealing with solutions and their colligative properties.
ewline ewline In the provided solution example, the calculation was used to find the molar mass of the compound Q after determining its van't Hoff factor and its composition by mass. To do this, we use the percentages by mass of each element in the compound and the known molar mass of these elements.
ewline ewline For more complex molecules, students might need to consider the contribution of each atom or ion to the total molar mass. Clarity in explaining the proportions of different ions in a solute, as well as how these proportions are used to calculate the molar mass of a compound from a given mass percent, can greatly aid in student comprehension. Incorporating this concept into exercises like the one discussed allows students to apply theoretical knowledge to practical problems, reinforcing their understanding of molar mass calculations.