91Ó°ÊÓ

Begin by converting the given mass of NaCl to moles. To do this, divide the mass of NaCl by its molar mass (58.44 g/mol): \[ \text{moles of NaCl} = \frac{3.4 \ \mathrm{g}}{58.44 \ \mathrm{g/mol}} = 0.0582 \ \mathrm{mol} \] Next, divide the moles of NaCl by the volume of the solution (1 L) to find the molarity (M): \[ \text{Molarity (M)} = \frac{0.0582 \ \mathrm{mol}}{1 \ \mathrm{L}} = 0.0582 \ \mathrm{M} \]

The van 't Hoff equation requires the temperature to be in Kelvin. To convert the given temperature (20°C) to Kelvin, just add 273.15: \[ T(K) = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \ \mathrm{K} \]

Now that we have the molarity and the temperature in Kelvin, we can use the van 't Hoff equation to find the osmotic pressure (Π). The equation is as follows: \[ \begin{aligned} \mathrm{Π} &= \mathrm{(i)} \times (\mathrm{M}) \times (\mathrm{R}) \times (\mathrm{T}) \end{aligned} \] where - i = the number of particles produced upon dissolving the solute (for NaCl, it is 2, as NaCl dissociates into Na+ and Cl- ions) - M = the molarity of the solute - R = the ideal gas constant (0.0821 L atm/mol K) - T = the temperature in Kelvin Plugging in the values, we get: \[ \begin{aligned} \mathrm{Π} &= 2 \times 0.0582 \ \mathrm{M} \times 0.0821 \ \mathrm{L\ atm/mol\ K} \times 293.15 \ \mathrm{K} \\ &= 2.81 \ \mathrm{atm} \end{aligned} \]

The osmotic pressure of seawater at \(20^{\circ} \mathrm{C}\), assuming the solute consists entirely of \(\mathrm{NaCl}\), is approximately 2.81 atm.