An aqueous solution is \(1.00 \%\) 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.

We are given the mass percentage of NaCl in the solution as \(1.00\%\) and the density of the solution as \(1.071\,g/cm^3\). First, let's assume a mass of 100 g of the solution, which contains 1 g of NaCl and 99 g of water. Now, we can convert the mass of NaCl to moles using its molar mass and the mass of water in kilograms: Number of moles of NaCl \(= \frac{1\,g}{58.44\,g/mol} = 0.0171\,mol\) Mass of water \(= \frac{99\,g}{1000\,g/kg} = 0.099\,kg\) Now, the molality of NaCl can be calculated by dividing the moles by the mass of water in kg: \(molality = \frac{0.0171\,mol}{0.099\,kg} = 0.173\,mol/kg\)

The observed osmotic pressure is given as \(7.83\,atm\). We can use the osmotic pressure equation to determine the effective concentration of ions in the solution: \(π = cRT\) where π is the osmotic pressure, c is the concentration in mol/L, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin (298 K). \(7.83\,atm = c (0.0821\,L\,atm/mol\,K) (298\,K)\) c \(= \frac{7.83\,atm}{(0.0821\,L\,atm/mol\,K)(298\,K)} = 0.321\,mol/L\) This calculated concentration is the effective concentration that considers the formation of ion pairs. We can use the molality of the NaCl solution to find the actual concentration of ions: Actual concentration of ions \(= 2 \times molality = 2 \times 0.173\,mol/kg = 0.346\,mol/L\) Now, we can calculate the fraction of ion pairs by comparing the actual and effective concentrations: Fraction of ion pairs \(= 1 - \frac{effective\ concentration}{actual\ concentration} = 1 - \frac{0.321\,mol/L}{0.346\,mol/L} = 0.072\) About 7.2% of the moles of NaCl in the solution exist as ion pairs.

Finally, we can use the freezing point depression equation to calculate the observed freezing point of the solution: ΔT_f \(= K_f \cdot m \cdot i\) where ΔT_f is the freezing point depression, K_f is the cryoscopic constant for water (1.853 K kg/mol), m is the molality of the NaCl solution (0.173 mol/kg), and i is the van't Hoff factor. In this case, consider i = 1.928 (two ions with 7.2% ion pairs) ΔT_f \(= (1.853\,K\,kg/mol) (0.173\,mol/kg) (1.928) = 0.617\,K\) To find the freezing point of the solution, subtract the freezing point depression from the normal freezing point of water (0 °C): Observed freezing point \(= 0\,°C - 0.617\,K = -0.617\,°C\) The observed freezing point of the solution is \(-0.617\,°C\).