Question

Consider an aqueous solution containing sodium chloride that has a density of \(1.01 \mathrm{~g} / \mathrm{mL}\). Assume the solution behaves ideally. The freezing point of this solution at \(1.0 \mathrm{~atm}\) is \(-1.28^{\circ} \mathrm{C}\). Calculate the percent composition of this solution (by mass).

Short answer

The percent composition of sodium chloride in the aqueous solution is approximately 0.99%.

Step-by-step solution

Identify known values and formula for freezing point depression

In this case, we are given the freezing point depression (∆t) of the sodium chloride solution, which is -1.28˚C. The freezing point of pure water at 1.0 atm is 0˚C. We can use the freezing point depression formula, \( \Delta T_f = K_f \times m \times i \) Where: - ∆Tf is the freezing point depression - Kf is the freezing point depression constant (1.86 °C·kg/mol for water) - m is the molality (moles of solute/kg of solvent) - i is the van't Hoff factor (2 for sodium chloride, as it dissociates into two ions, Na+ and Cl-) We will now rearrange the formula to solve for molality.

Calculate the molality of the solution

Solving for molality using the formula above, we get: \( m = \frac{\Delta T_f}{K_f \times i} \) Plugging in the given values: \( m = \frac{-1.28}{1.86 \times 2} = -0.3441 \, \text{mol/kg} \) Since molality is a positive value, we can drop the negative sign: m = 0.3441 mol/kg

Calculate the mass of the sodium chloride in the solution

Now we have the molality (0.3441 mol/kg) of the solution. In order to find the mass of sodium chloride, we need to multiply the molality by the molar mass of sodium chloride (NaCl) and the mass of the solvent (water). To find the mass of the solvent (water), we first need to consider a known volume. We will use 1 liter (or 1000 ml) of solution as the known volume and use the given density to find the mass. Density = Mass / Volume Mass of solution = Density × Volume = 1.01 g/mL × 1000 mL = 1010 g We know that 1000 ml of water weighs 1000 g. Since the density of the solution is 1.01 g/mL, there is an additional mass of 10 g (i.e., 1010 g - 1000 g = 10 g). This additional mass is due to the sodium chloride. Now, we can calculate the percent composition of sodium chloride in the solution:

Calculate the percent composition of sodium chloride in the solution

Percent composition can be calculated using the formula: Percent composition = (Mass of solute / Mass of solution) × 100 Plugging in the values: Percent composition = (10 g / 1010 g) × 100 = 0.9901% Therefore, the percent composition of the sodium chloride in the solution is approximately 0.99%.

Key concepts

Molality

Molality is a way to express the concentration of a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality depends on the mass of the solvent.

• To calculate molality, you can use the formula: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
• In scenarios involving freezing point depression, molality is particularly useful since it is independent of temperature changes, making it more reliable than molarity in temperature-variable conditions.

For example, if you have 0.3441 moles of sodium chloride dissolved in 1 kg of water, the molality is expressed as 0.3441 mol/kg. This tells you that for every kilogram of water, there are 0.3441 moles of solute present.

van't Hoff Factor

The van't Hoff factor (\(i\)) is a measure of the effect of solute particles on the colligative properties of a solution. It represents the number of particles that a compound forms when it dissolves in solution. For ionic compounds like sodium chloride (NaCl), it accounts for dissociation into ions.

• For NaCl, the van't Hoff factor is 2 because it dissociates into two ions: Na\(^+\) and Cl\(^-\).
• The formula for adjusting molality due to dissociation is given by: \[ m_{eff} = m \times i \]
• This relationship explains why saltwater has a lower freezing point than pure water. The more ions present, the greater the impact on the freezing point.

Essentially, for every unit of NaCl that dissolves, two particles form, leading to a more significant freezing point depression. Knowing the van't Hoff factor helps us calculate how much the presence of solute affects the freezing point or boiling point of a solution.

Percent Composition

Percent composition is a way to express how much of a component is present in a solution by mass. It helps in understanding the ratio of solute to the total solution. Let's look at how to calculate it:

• The formula is:\[ \text{Percent Composition} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 \]
• This approach is beneficial in chemistry for formulating and understanding solutions, especially in determining substance concentration.

In the given example, the solution's density helps find the total mass of the solution. The additional mass in the solution beyond the pure solvent is attributed to the sodium chloride. By calculating the mass of the solute (NaCl) and dividing it by the total mass of the solution, then multiplying by 100, you get the percent composition. This method lets you evaluate exactly how much sodium chloride is present, in this case, around 0.99%.

Sodium Chloride Solution

A sodium chloride solution is a mixture of salt (NaCl) in water. It's often referred to as saltwater and is known for its many practical uses, including in cooking, medical treatments, and chemical reactions. This type of solution is a classic example in chemistry to study colligative properties, such as boiling and freezing point changes.

• The preparation of a sodium chloride solution involves dissolving NaCl in water, and its concentration can be quantified using measurements like molality or percent composition.
• When forming a solution, it's important to understand how properties like the van't Hoff factor affect its behavior. For instance, NaCl dissociates into ions, which is critical in colligative property calculations.

This solution perfectly illustrates concepts like molality and freezing point depression. The practical implications of these solutions are vast, from de-icing roads in winter to preserving food.