In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water contains about \(560 \mu \mathrm{g}\) of dissolved solids per milliliter. Reverse osmosis units are marketed in this area to soften water. A typical unit exerts a pressure of 8.0 atm and can produce 45 L water per day. a. Assuming all of the dissolved solids are \(\mathrm{MgCO}_{3}\) and assuming a temperature of \(27^{\circ} \mathrm{C},\) what total volume of water must be processed to produce 45 L pure water? b. Would the same system work for purifying seawater? (Assume seawater is 0.60 \(M\) NaCl.)

First, we need to find out how many grams of \(\mathrm{MgCO}_{3}\) are present in 45 L of water. Since the tap water contains 560 μg of dissolved solids per milliliter, we can use this information to calculate the total dissolved \(\mathrm{MgCO}_{3}\). Total dissolved solids in 45 L of water = 560 μg/mL \(\times\) 1000 mL/L \(\times\) 45 L = \(25.2 \times 10^{3}\) μg = 25.2 g

Next, we need to find out the number of moles of \(\mathrm{MgCO}_{3}\) present in 25.2 g. For this, we will use the molar mass of \(\mathrm{MgCO}_{3}\), which is approximately 84.31 g/mol. Number of moles of \(\mathrm{MgCO}_{3}\) = \(\dfrac{25.2 \text{g}}{84.31 \text{g/mol}} = 0.299 \space \text{mol}\)

Now we'll calculate the osmotic pressure, which is the pressure required to stop the net flow of water due to a concentration gradient. For this, we'll use the formula: Osmotic pressure (Π) = nRT/V Here, n is the number of moles of solute, R is the ideal gas constant (\(0.0821 \;\mathrm{L\,atm\,K^{-1}\,mol^{-1}}\)), T is the temperature in Kelvin (273 + 27 = 300 K), and V is the volume of water in liters. Osmotic pressure (Π) = \(\dfrac{(0.299 \mathrm{\, mol})(0.0821 \mathrm{\, L\, atm\, K^{-1}\,mol^{-1}})(300 \mathrm{\, K})}{V}\) Using the information provided in the problem, the reverse osmosis unit exerts a pressure of 8.0 atm. So, we'll set the osmotic pressure equal to 8.0 atm and solve for V: 8.0 atm = \(\dfrac{(0.299 \mathrm{\, mol})(0.0821 \mathrm{\, L\, atm\, K^{-1}\,mol^{-1}})(300 \mathrm{\, K})}{V}\)

To find the total volume of water processed, we'll solve for V in the above equation: V = \(\dfrac{(0.299 \mathrm{\, mol})(0.0821 \mathrm{\, L\, atm\, K^{-1}\,mol^{-1}})(300 \mathrm{\, K})}{8.0 \mathrm{\, atm}} = 91.4\,\mathrm{L}\) So, the system must process a total of 91.4 L of water to produce 45 L of pure water.

To check if the system works for purifying seawater (assuming 0.60 M NaCl), we need to calculate the osmotic pressure required to remove NaCl. Osmotic pressure (Π) = MRT Here, M is the molarity of the solute (0.60 M), R is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹), and T is the temperature in Kelvin (300 K). Osmotic pressure (Π) = \((0.60 \,\mathrm{M})(0.0821 \mathrm{\, L\, atm\, K^{-1}\,mol^{-1}})(300 \mathrm{\, K}) = 14.8\,\mathrm{atm}\) Since the osmotic pressure required for purifying seawater (14.8 atm) is greater than the pressure applied by the reverse osmosis unit (8.0 atm), the system would not work for purifying seawater.