Problem 64 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A quantity of \(3.125 \mathrm{~g}\) of a mixture of \(\mathrm{KCl}\) and \(\mathrm{NaCl}\) dissolved in \(1 \mathrm{~kg}\) of water produces a depression of \(0.186^{\circ} \mathrm{C}\) in freezing point. The molar ratio of \(\mathrm{KCl}\) to \(\mathrm{NaCl}\) in the solution (assuming complete dissociation of the salts) is \(\left(K_{f}=1.86\right.\) deg \(/\) molal \()\) (a) \(1: 3\) (b) \(2: 3\) (c) \(1: 1\) (d) \(3: 1\)

Step-by-Step Solution

Answer

Based on the calculations, the molar ratio of \(\mathrm{KCl}\) to \(\mathrm{NaCl}\) is option (b) 2:3.

Step 1: Determine the combined molality of the salts

Use the depression of freezing point formula, \(\Delta T_f = i \times K_f \times m\), where \(\Delta T_f\) is the depression of freeing point, \(i\) is the van't Hoff factor (which is the number of particles a compound dissociates into), \(K_f\) is the cryoscopic constant, and \(m\) is the molality. Given that \(\Delta T_f = 0.186^\circ \mathrm{C}\) and \(K_f = 1.86\) deg/molal, solve for \(m\).

Step 2: Calculate the moles of \(\mathrm{KCl}\) and \(\mathrm{NaCl}\) present

Since \(\mathrm{KCl}\) dissociates into 2 ions (\(\mathrm{K}^+\) and \(\mathrm{Cl}^-\)) and \(\mathrm{NaCl}\) also dissociates into 2 ions (\(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\)), the van't Hoff factor \(i\) for both salts is 2. Let \(x\) be the molality of \(\mathrm{KCl}\) and \(y\) be the molality of \(\mathrm{NaCl}\) in the solution. We can write the total molality as \(m = 2x + 2y\).

Step 3: Translate mass of the mixture into molar quantities

Convert the total mass of salts into moles using their molar masses (\(M_{\mathrm{KCl}} = 74.55 \mathrm{~g/mol}\) and \(M_{\mathrm{NaCl}} = 58.44 \mathrm{~g/mol}\)). Let \(n_{\mathrm{KCl}}\) be the moles of \(\mathrm{KCl}\) and \(n_{\mathrm{NaCl}}\) be the moles of \(\mathrm{NaCl}\) in the 3.125 g mixture. We can write the expressions for the moles as \(n_{\mathrm{KCl}} = \frac{3.125g \times f}{74.55gmol^{-1}}\) and \(n_{\mathrm{NaCl}} = \frac{3.125g \times (1-f)}{58.44 g mol^{-1}}\), where \(f\) is the fraction of \(\mathrm{KCl}\) in the mixture.

Step 4: Establish a relationship using molality and moles

The molality of each salt is the number of moles of the salt divided by the mass of the solvent in kilograms. For the system, \(x = \frac{n_{\mathrm{KCl}}}{1kg}\) and \(y = \frac{n_{\mathrm{NaCl}}}{1kg}\). We have two equations to solve for the two unknowns \(x\) and \(y\).

Step 5: Solve the system of equations

By combining the equations from Step 2 and Step 3, solve for \(x\) and \(y\). Substitute \(x\) and \(y\) back into the equation for \(m\) from Step 1 and solve for the values of \(f\).

Step 6: Calculate the molar ratio and choose the right option

Using the values of \(x\) and \(y\), calculate the molar ratio of \(\mathrm{KCl}\) to \(\mathrm{NaCl}\). The molar ratio is given by \(\frac{n_{\mathrm{KCl}}}{n_{\mathrm{NaCl}}} = \frac{x}{y}\). Match this ratio with the given options to find the correct answer.

Key Concepts

Freezing Point Depression

Freezing point depression is a colligative property that occurs when a solute is dissolved in a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. This happens because the solute particles disrupt the formation of the solid structure of the solvent, requiring a lower temperature to achieve the same structural integrity.

For instance, adding salt to ice lowers its freezing point, which is why salt is used to melt ice on roads during winter. In chemistry, the magnitude of freezing point depression can be calculated using the formula: \[ \Delta T_f = i \times K_f \times m \]where \(\Delta T_f\) is the change in freezing point, \(i\) is the van't Hoff factor (which indicates the number of particles the solute dissociates into in solution), \(K_f\) is the cryoscopic constant specific to the solvent, and \(m\) is the molality of the solution (moles of solute per kilogram of solvent).

Molality Calculation

Molality is a measure of the mole concentration of a solute in a solution. Unlike molarity, which is temperature-dependent, molality is not, because it is based on the mass of the solvent, not the volume of the solution. This makes molality particularly useful in situations involving temperature changes, such as freezing point depression.

To calculate molality, use the formula:\[ m = \frac{{moles \ of \ solute}}{{kilograms \ of \ solvent}} \]The key to accurately determining molality is knowing the amount of solute and the mass of the solvent. Precision in these measurements will ensure correct calculation of colligative properties.

van't Hoff Factor

The van't Hoff factor, symbolized as \(i\), is integral to understanding colligative properties. It represents the number of particles into which a compound dissociates when dissolved in a solvent. For non-electrolytes that do not dissociate in solution, \(i\) is typically 1. For electrolytes like salts, \(i\) is equal to the total number of ions produced.

In the case of complete dissociation of a salt like sodium chloride (\(NaCl\)), which dissociates into two ions, \(Na^+\) and \(Cl^-\), the van't Hoff factor is 2. Accurately predicting \(i\) is crucial when applying the freezing point depression formula, as it affects the extent to which the solute impacts the solvent's properties.

Dissociation of Salts

Dissociation of salts is the process by which an ionic compound separates into its constituent ions in solution. This phenomenon is particularly important when studying colligative properties since the degree of dissociation affects the values of these properties. Salts like \(KCl\) and \(NaCl\) typically dissociate completely in water, producing \(K^+\), \(Na^+\), and \(Cl^-\) ions.

Understanding this process is essential for solving problems involving changes in boiling point, freezing point, or vapor pressure. When salts dissociate, the total number of solute particles in the solution increases, which is reflected by the van't Hoff factor in certain colligative property equations.