Problem 24 - Chemistry: An Atoms-Focused Approach

Question

Calculate the osmotic pressure of each of the following aqueous solutions at \(27^{\circ} \mathrm{C}\) a. \(10.0 \mathrm{g}\) of \(\mathrm{NaCl}\) in \(1.50 \mathrm{L}\) of solution b. \(10.0 \mathrm{mg} / \mathrm{L}\) of \(\mathrm{LiNO}_{3}\) c. \(0.222 M\) glucose d. \(0.00764 M \mathrm{K}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

Question: Calculate the osmotic pressure for each of the following aqueous solutions at 27°C: a) 10.0 g NaCl in 1.50 L of solution, b) 10.0 mg/L LiNO₃, c) 0.222 M glucose, and d) 0.00764 M K₂SO₄. Answer: a) The osmotic pressure for the 10.0 g NaCl in 1.50 L of solution is approximately 5.60 atm. b) The osmotic pressure for the 10.0 mg/L LiNO₃ solution is approximately 0.00714 atm. c) The osmotic pressure for the 0.222 M glucose solution is approximately 5.46 atm. d) The osmotic pressure for the 0.00764 M K₂SO₄ solution is approximately 0.562 atm.

Step 1: a. 10.0 g NaCl in 1.50 L of solution

First, determine the number of moles of NaCl and find the molarity. To calculate moles of NaCl: moles = mass (g) / molar mass (g/mol) NaCl molar mass: 58.44 g/mol moles = 10.0 g / 58.44 g/mol ≈ 0.1712 mol To calculate molarity (M): M = moles / volume M = 0.1712 mol / 1.50 L ≈ 0.1141 M For NaCl, the value of van 't Hoff factor (i) is 2, since it dissociates into two ions (Na\(^+\) and Cl\(^−\)). Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT R = 0.0821 L atm/mol K T = 27°C + 273.15 = 300.15 K Osmotic Pressure = 2 × 0.1141 M × 0.0821 L atm/mol K × 300.15 K ≈ 5.60 atm

Step 2: b. 10.0 mg/L LiNO₃

First, determine the number of moles of LiNO₃ and find the molarity. To calculate moles of LiNO₃: moles = mass (g) / molar mass (g/mol) LiNO₃ molar mass: 68.94 g/mol (approximately) moles = 10.0 mg × 10\(^{-3}\) / 68.94 g/mol ≈ 1.45 × 10\(^{-4}\) mol To calculate molarity (M): M = moles / volume M = 1.45 × 10\(^{-4}\) mol / 1 L = 1.45 × 10\(^{-4}\) M For LiNO₃, the value of van 't Hoff factor (i) is 2, since it dissociates into two ions (Li\(^+\) and NO₃\(^−\)). Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT Osmotic Pressure = 2 × 1.45 × 10\(^{-4}\) M × 0.0821 L atm/mol K × 300.15 K ≈ 0.00714 atm

Step 3: c. 0.222 M glucose

Glucose (C₆H₁₂O₆) is a non-electrolyte and does not dissociate in aqueous solution, so the van 't Hoff factor (i) for glucose is 1. Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT Osmotic Pressure = 1 × 0.222 M × 0.0821 L atm/mol K × 300.15 K ≈ 5.46 atm

Step 4: d. 0.00764 M K₂SO₄

First, determine the van 't Hoff factor (i) for K₂SO₄. K₂SO₄ dissociates into 3 ions: 2K\(^+\) and 1SO\(_4^{2-}\), so the van 't Hoff factor (i) is 3. Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT Osmotic Pressure = 3 × 0.00764 M × 0.0821 L atm/mol K × 300.15 K ≈ 0.562 atm