How would you prepare 1.0 L of an aqueous solution of sodium chloride having an osmotic pressure of 15 atm at \(22^{\circ} \mathrm{C} ?\) Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

: The van't Hoff equation relates osmotic pressure (Π) to the concentration of the solute (c), and the ideal gas constant (R) and the temperature (T). The equation is given by: \[Π = icRT\] Where: i: van't Hoff factor, the number of particles produced by the solute in the solution c: concentration of the solute (in mol/L) R: ideal gas constant (0.0821 L atm/mol K) T: temperature (in Kelvin) We know that sodium chloride completely dissociates into Na+ and Cl- ions in solution, so the van't Hoff factor (i) is 2. We are given the osmotic pressure (Π) as 15 atm and temperature as 22°C; therefore, we need to convert this to Kelvin: T(K) = T(°C) + 273.15 T(K) = 22 + 273.15 T(K) = 295.15 K Now, we can solve for the concentration (c): 15 atm = 2c × 0.0821 L atm/mol K × 295.15 K

: Now let's solve the equation to find the concentration (c): \[c = \dfrac{15\,\text{atm}}{2 \times 0.0821\, \text{L atm/mol K} \times 295.15\,\text{K}}\] \[c \approx 1.548\,\text{mol/L}\]

: We need to prepare 1.0 L of the aqueous solution, so we can calculate the total amount of sodium chloride needed for this concentration: Amount (mol) = Concentration × Volume Amount (mol) = 1.548 mol/L × 1.0 L Amount (mol) = 1.548 mol

: Finally, we will convert the amount of sodium chloride in moles to grams using its molar mass: Molar mass of sodium chloride = 58.44 g/mol Amount (g) = Amount(mol) × Molar mass Amount (g) = 1.548 mol × 58.44 g/mol Amount (g) ≈ 90.41 g

: To prepare 1.0 L of an aqueous solution of sodium chloride with an osmotic pressure of 15 atm at 22 °C, dissolve approximately 90.41 g of sodium chloride in enough water to make the total volume 1.0 L. That's it! The solution is ready to use.