University:
College
Course:
College Chemistry
Academic year:

2022
Views:

302

Pages:

13
Author:

StellarSymphony

Worksheet: Solutions and Colligative Properties Set A: 1) [Al³⁺] = (0.115 mol AlCl₃ / 0.820 L) x (1 mol Al³⁺ / 1 mol AlCl₃) = 0.201 M Al³⁺ [Cl⁻] = (0.115 mol AlCl₃ / 0.820 L) x (3 mol Cl⁻ / 1 mol AlCl₃) = 0.603 M Cl⁻ 2) M(H⁺) = (0.027 L x 0.25 mol HNO₃ / L) x (1 mol H⁺ / 1 mol HNO₃) = 0.011 M H⁺ M(Ca²⁺) = (0.036 L x 0.42 mol Ca(NO₃)₂ / L) x (1 mol Ca²⁺ / 1 mol Ca(NO₃)₂) = 0.024 M Ca²⁺ NO₃⁻ mol from HNO₃ = 0.027 L x 0.25 mol HNO₃ / L x (1 mol NO₃⁻ / 1 mol HNO₃) = 6.8 x 10⁻³ mol NO₃⁻ NO₃⁻ mol from Ca(NO₃)₂ = 0.036 L x 0.42 mol Ca(NO₃)₂ / L x (2 mol NO₃⁻ / 1 mol Ca(NO₃)₂) = 3.0 x 10⁻² mol NO₃⁻ M(NO₃⁻) = (6.8 x 10⁻³ + 3.0 x 10⁻²) mol NO₃⁻ / (0.027 L + 0.036 L) = 0.58 M NO₃⁻ Set A 3) mol K₂SO₄ 0.035 L x 0.12 mol K₂SO₄/L = 0.015 mol K₂SO₄ mol K₃PO₄ 0.027 L x 0.17 mol K₃PO₄/L = 4.6 x 10⁻³ mol K₃PO₄ M SO₄²⁻ = (0.015 mol K₂SO₄ x 1 mol SO₄²⁻ / 1 mol K₂SO₄) / (0.035 L + 0.027 L) = 0.24 M SO₄²⁻ M PO₄³⁻ = (4.6 x 10⁻³ mol K₃PO₄ x 1 mol PO₄³⁻ / 1 mol K₃PO₄) / (0.035 L + 0.027 L) = 0.074 M PO₄³⁻ M K⁺ = ((0.015 mol K₂SO₄ x 2 mol K⁺ / 1 mol K₂SO₄) + (4.6 x 10⁻³ mol K₃PO₄ x 3 mol K⁺ / 1 mol K₃PO₄)) / (0.035 L + 0.027 L) = 0.71 M K⁺ 3) moles heptane: 5.32g C7H16 / 100.1 g/mol C7H16 = 0.0531 mol C7H16 moles octane: 6.09g C8H18 / 114.2 g/mol C8H18 = 0.0531 mol C8H18 Total Vapor Pressure = Xheptane * P°heptane + Xoctane * P°octane Total Vapor Pressure = 0.0531 mol C7H16 / (0.0531 mol C7H16 + 0.0531 mol C8H18) * 2.13 atm + 0.0531 mol C8H18 / (0.0531 mol C7H16 + 0.0531 mol C8H18) * 1.41 atm Total Vapor Pressure = 0.5 * 2.13 atm + 0.5 * 1.41 atm = 1.77 atm 4) ΔT = i * Kf * m_compd m_compd = 0.0400 m_compd mol compd = 0.0400 m_compd * 0.0800 kg benzene = 0.0032 mol compd Molar mass (compd) = 0.800 g compd / 0.0032 mol compd = 250 g/mol Empirical formula mass = (4 x 12.01 g/mol C) + (2 x 1.01 g/mol H) + 14.01 g/mol N = 64 g/mol n = Molar mass (compd) / Empirical formula mass = 250 g/mol / 64 g/mol ≈ 4 ∴ 4 x C₄H₂N = C₈H₈N₄ 5) P_total = P_A + P_B P_A = X_A * P°_A P_B = X_B * P°_B P_A at Normal Boiling Point = 1 atm = 0.650 mol A / (0.650 mol A + 2.70 mol B) * (P°_A) P°_B = 1.77 atm 6) π = i M R T 12.7 torr / (760 torr/atm) = (1) M_protein (0.021 L/L) (273.15 + 25) K M_protein = 6.83 x 10⁻⁴ mol Protein / L solution moles protein = 0.900 L soln x (6.83 x 10⁻⁴ mol Protein / L soln) = 6.15 x 10⁻⁴ mol Protein Molar mass protein = 30.0 g Protein / 6.15 x 10⁻⁴ mol Protein = 4.88 x 10⁴ g/mol Protein