1. **Problem 5: Sulfuric Acid Solution Calculations** Given: mass of H₂SO₄ = 55 g, mass of water = 350 g, density of H₂SO₄ = 1.84 g/mL Molar mass of H₂SO₄ = 98.079 g/mol Molar mass of H₂O = 18.015 g/mol --- **a. Mole fraction of solute (H₂SO₄) and solvent (H₂O):** 1. Calculate moles of H₂SO₄: $$n_{H_2SO_4} = \frac{55}{98.079} \approx 0.5605\, mol$$ 2. Calculate moles of H₂O: $$n_{H_2O} = \frac{350}{18.015} \approx 19.43\, mol$$ 3. Calculate mole fractions: $$X_{H_2SO_4} = \frac{0.5605}{0.5605 + 19.43} \approx 0.028\quad\text{and}\quad X_{H_2O} = 1 - X_{H_2SO_4} = 0.972$$ --- **b. Normality of the solution:** 1. Volume of H₂SO₄: $$V = \frac{mass}{density} = \frac{55}{1.84} \approx 29.89\, mL = 0.02989\, L$$ 2. Moles of H₂SO₄ = 0.5605 mol 3. Normality (N) = equivalents per liter. H₂SO₄ has 2 equivalents per mole (2 H⁺ ions): $$N = \frac{2 \times 0.5605}{0.02989} \approx 37.5\, N$$ --- **c. Molality (m):** Molality = moles of solute per kg of solvent: $$m = \frac{0.5605}{0.350} \approx 1.60\, mol/kg$$ --- **d. Molarity (M):** Total volume = volume of H₂SO₄ + volume of water (approximate volume of water = mass/density of water = 350 g / 1 g/mL = 350 mL) $$V_{total} = 29.89 + 350 = 379.89\, mL = 0.3799\, L$$ Molarity: $$M = \frac{0.5605}{0.3799} \approx 1.48\, M$$ --- **e. Volume-volume percent of solute to solvent:** $$\text{Volume %} = \frac{V_{solute}}{V_{solvent}} \times 100 = \frac{29.89}{350} \times 100 \approx 8.54\%$$ --- 2. **Problem 6: Urea Solution Calculations** Given: mass urea = 40.0 g, mass water = 160 g, temperature = 35°C, vapor pressure of pure water = 43.76 torr, density of urea = 1.32 g/cm³ Molar mass of urea (CH₄N₂O) = 60.06 g/mol --- **a. Vapor pressure of the solution:** 1. Calculate moles of urea: $$n_{urea} = \frac{40.0}{60.06} \approx 0.666\, mol$$ 2. Calculate moles of water: $$n_{H_2O} = \frac{160}{18.015} \approx 8.88\, mol$$ 3. Mole fraction of water: $$X_{H_2O} = \frac{8.88}{8.88 + 0.666} \approx 0.930$$ 4. Vapor pressure of solution: $$P_{solution} = X_{H_2O} \times P^0_{H_2O} = 0.930 \times 43.76 \approx 40.7\, torr$$ --- **b. Vapor Pressure Lowering:** $$\Delta P = P^0_{H_2O} - P_{solution} = 43.76 - 40.7 = 3.06\, torr$$ --- **c. Molarity of the solution:** 1. Volume of urea: $$V = \frac{mass}{density} = \frac{40.0}{1.32} \approx 30.3\, mL = 0.0303\, L$$ 2. Volume of water (approx): $$V_{H_2O} = \frac{160}{1} = 160\, mL = 0.160\, L$$ 3. Total volume: $$V_{total} = 0.0303 + 0.160 = 0.1903\, L$$ 4. Molarity: $$M = \frac{0.666}{0.1903} \approx 3.50\, M$$ --- **d. Osmotic Pressure ($\Pi$):** Use formula: $$\Pi = MRT$$ Where $R = 0.0821\, L\cdot atm/(mol\cdot K)$, $T = 35 + 273 = 308 K$ $$\Pi = 3.50 \times 0.0821 \times 308 \approx 88.4\, atm$$ --- **e. Freezing point depression:** 1. Molality: $$m = \frac{0.666}{0.160} = 4.16\, mol/kg$$ 2. Freezing point depression constant for water $K_f = 1.86\, ^\circ C/m$ 3. Depression: $$\Delta T_f = K_f \times m = 1.86 \times 4.16 = 7.74\, ^\circ C$$ 4. Freezing point of solution: $$T_f = 0 - 7.74 = -7.74\, ^\circ C$$