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).

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.

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

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:

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%.