Question

An aqueous solution is 1.00\(\% \mathrm{NaCl}\) by mass and has a density of 1.071 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) . The observed osmotic pressure of this solution is 7.83 atm at \(25^{\circ} \mathrm{C}\) . a. What fraction of the moles of NaCl in this solution exist as ion pairs? b. Calculate the freezing point that would be observed for this solution.

Short answer

The fraction of ion pairs in the NaCl solution is approximately \(1 - \frac{effective \, molality}{molality}\). The observed freezing point of the solution can be calculated using the formula ΔT_f = K_f × molality × i and is found by subtracting the freezing point depression from the normal freezing point of water (0°C).

Step-by-step solution

Calculate the molality of the NaCl solution

We first need to find the molality (moles of solute/kg of solvent) of the NaCl solution. The solution is given as 1.00% NaCl by mass. So, we have: 1.00 g NaCl per 100 g of solution Using the density of the solution, we can find the mass of water present in the solution: density = mass of solution / volume of solution Rearranging, we have mass of the solution = density × volume of the solution. Since we're dealing with 100 g of solution, we can find the volume as follows: Volume of the solution = mass of solution / density \(Volume_{solution} = \frac{100 \, g}{1.071 \, \frac{g}{cm^3}} \) Now, subtract the mass of NaCl to obtain the mass of water: Mass of water = mass of the solution - mass of NaCl = 100 g - 1.00 g Next, calculate the number of moles of NaCl: Number of moles of NaCl = mass of NaCl / molar mass of NaCl. Now, we can find the molality of the NaCl solution: Molality_k = (moles of NaCl)/(mass of water in kg).

Use osmotic pressure to find effective molality

Using the given osmotic pressure (7.83 atm), we can calculate the effective molality of the solution. The osmotic pressure can be described as: \(Π = i \cdot n \cdot R \cdot T \cdot Φ\), where Π is the osmotic pressure, i is the van 't Hoff factor, n is the molality, R is the gas constant, T is the temperature in K and Φ is the osmotic coefficient. Given that: the osmotic pressure Π = 7.83 atm, R = 0.0821 L*atm/mol*K, and T = 298 K (25°C + 273), we can rearrange the equation to find the effective molality of the NaCl solution. Effective molality = \( i \cdot n \cdot Φ\)

Calculate the fraction of ion pairs

Now that we have the molality and effective molality of the NaCl solution, we can use them to find the fraction of ion-pairs in the solution: Fraction of ion-pairs = \(1 - \frac{effective \, molality}{molality}\)

Calculate the freezing point depression

To calculate the freezing point depression, we will use the formula: ΔT_f = K_f × molality × i, where ΔT_f is the freezing point depression, K_f is the cryoscopic constant for water (1.86 °C kg/mol), molality is calculated as in Step 1, and i is the van't Hoff factor. We know that for NaCl, ideally, i=2 in a dilute solution, but here we have ion-pairs, so the effective i should be adjusted according to the fraction obtained in Step 3. Now, calculate the observed freezing point depression using the adjusted i value. Finally, subtract the obtained value from the normal freezing point of water (0°C) to find the freezing point of the solution: Freezing point of the solution = normal freezing point of water - freezing point depression.

Key concepts

Molality

Molality is a way to express the concentration of a solution, and it plays a crucial role in determining colligative properties. To calculate molality, you need to know the moles of solute and the mass of the solvent in kilograms. Here's a simple breakdown:

• Take the mass of your solute in grams and convert it to moles by dividing it by the solute's molar mass.
• Measure the mass of the solvent in the solution, usually in kilograms.
• Divide the number of moles of solute by the mass of the solvent in kilograms to get molality.

In the context of the NaCl aqueous solution question, molality allows us to understand how much solute is present in a given mass of solvent. For this particular solution, because the mass percentages and density are given, you first find how much of the solution is NaCl and how much is water (the solvent). From there, you compute the molality, which is essential for understanding other phenomena like osmotic pressure and freezing point depression.

Freezing Point Depression

Freezing point depression is a colligative property, which means it depends on the number of solute particles in a solution rather than their identity. This property signifies how the presence of solute lowers the freezing point of the solvent. Here's how it works:

• When solute particles are added to a solvent, they disrupt the solvent’s normal freezing pattern.
• The more solute particles present, the lower the temperature required for the solvent molecules to arrange into a solid structure.
• The freezing point depression (\(\Delta T_f\)) can be calculated using the formula: \(\Delta T_f = K_f \times \text{molality} \times i\), where \(K_f\) is the cryoscopic constant (specific for each solvent), while \(i\) is the van 't Hoff factor for the solute.

In this exercise, you would calculate the freezing point depression by using the molality you previously calculated and taking into account the effective van't Hoff factor derived from the number of ion pairs. This effective \(i\) acknowledges that some solute ions form pairs and do not contribute as separate entities to the freezing point depression.

Ion Pairs

In solutions like the NaCl aqueous mix discussed, not all dissolved units act as separate ions. Some form ion pairs. Ion pairs are when a positive ion (like \(\text{Na}^+\)) binds closely with a negative ion (like \(\text{Cl}^-\)), forming a pair that behaves like one particle in a solution.Understanding ion pairs helps explain why calculated colligative properties sometimes deviate from expected theorems. In ideal solutions, NaCl would completely dissociate, giving an \(i\) value of 2 because each formula unit splits into two ions. However, ion pairs reduce the effective number of particles, thereby lowering the effective \(i\), leading to actual values of freezing point depression or osmotic pressure being less than the theoretical prediction.

• To find the fraction of ion pairs, compare the effective molality (from osmotic pressure) with the theoretical molality.
• The lower effective molality indicates more ion pairs are present.
• This fraction is crucial for accurately predicting real-life behavior of solutions in various chemical processes.

Understanding ion pairs is invaluable for chemists working with solutions in both practical and theoretical settings.