University:
College
Course:
College Chemistry
Academic year:

2022
Views:

131

Pages:

13
Author:

SpaceCadet101

Worksheet Solutions and Colligative Properties Chem 111 SET A: 1) Find the molarity of all ions in a solution that contains 0.165 moles of aluminum chloride in 820 mL solution. [Al³⁺] = 0.165 mol AlCl₃ / 0.820 L = 0.201 M [Cl⁻] = 0.165 mol AlCl₃ * (3 mol Cl⁻ / 1 mol AlCl₃) / 0.820 L = 0.603 M 2) Find the molarity of each ion present after mixing 27 mL of 0.25 M HNO₃ with 36 mL of 0.42 M Ca(NO₃)₂. (Note: There is no reaction taking place.) Answer: [H⁺] = 0.11 M, [NO₃⁻] = 0.58 M, [Ca²⁺] = 0.24 M 1) Find moles H⁺: mol H⁺ = M HNO₃ * V HNO₃ = 0.25 mol/L * 0.027 L = 6.8 × 10⁻³ mol H⁺ 2) Find moles Ca²⁺: mol Ca²⁺ = M Ca(NO₃)₂ * V Ca(NO₃)₂ = 0.42 mol/L * 0.036 L = 1.5 × 10⁻² mol Ca²⁺ [H⁺] = (6.8 × 10⁻³ mol H⁺) / 0.063 L = 0.11 M [Ca²⁺] = (1.5 × 10⁻² mol Ca²⁺) / 0.063 L = 0.24 M moles NO₃⁻ from HNO₃: 6.8 × 10⁻³ mol HNO₃ * (1 mol NO₃⁻ / 1 mol HNO₃) = 6.8 × 10⁻³ mol NO₃⁻ moles NO₃⁻ from Ca(NO₃)₂: 1.5 × 10⁻² mol Ca(NO₃)₂ * (2 mol NO₃⁻ / 1 mol Ca(NO₃)₂) = 3.0 × 10⁻² mol NO₃⁻ [NO₃⁻] = (6.8 × 10⁻³ + 3.0 × 10⁻²) mol NO₃⁻ / 0.063 L = 0.58 M 3) Find the molarity of each ion present after mixing 35 mL of 0.42 M K₂SO₄ with 27 mL of 0.17 M K₃PO₄. Answer: [K⁺] = 0.71 M, [SO₄²⁻] = 0.24 M, [PO₄³⁻] = 0.074 M. Find moles of K₂SO₄: M K₂SO₄ * V K₂SO₄ = 0.42 mol/L * 0.035 L = 0.015 mol K₂SO₄ Find moles of K₃PO₄: M K₃PO₄ * V K₃PO₄ = 0.17 mol/L * 0.027 L = 4.6 × 10⁻³ mol K₃PO₄ Find conc. of SO₄²⁻: 0.015 mol K₂SO₄ * (1 mol SO₄²⁻ / 1 mol K₂SO₄) = 0.015 mol SO₄²⁻ [SO₄²⁻] = 0.015 mol / 0.062 L = 0.24 M Find conc. of K⁺: 0.015 mol K₂SO₄ * (2 mol K⁺ / 1 mol K₂SO₄) = 0.030 mol K⁺ 4.6 × 10⁻³ mol K₃PO₄ * (3 mol K⁺ / 1 mol K₃PO₄) = 0.014 mol K⁺ [K⁺] = (0.030 + 0.014) mol K⁺ / 0.062 L = 0.71 M Find conc. of PO₄³⁻: 4.6 × 10⁻³ mol K₃PO₄ * (1 mol PO₄³⁻ / 1 mol K₃PO₄) = 0.0046 mol PO₄³⁻ [PO₄³⁻] = 0.0046 mol / 0.062 L = 0.074 M 3) Heptane, C₇H₁₆, and octane, C₈H₁₈, form ideal solutions. At 40 °C, the vapor pressure of pure heptane is 0.522 atm, and the vapor pressure of pure octane is 0.238 atm. A solution is made of 5.32 g heptane and 8.80 g octane. Calculate the mole fraction of octane in the vapor at the above temperature. Total vapor pressure = X_heptane * P°_heptane + X_octane * P°_octane = (5.32 g / 100.1 g/mol) / (5.32 g / 100.1 g/mol + 8.80 g / 114.2 g/mol) * 0.522 atm + (8.80 g / 114.2 g/mol) / (5.32 g / 100.1 g/mol + 8.80 g / 114.2 g/mol) * 0.238 atm = 0.213 atm + 0.141 atm = 0.354 atm X_octane in vapor = 0.141 atm / 0.354 atm = 0.398 4) What is the molar mass and molecular formula of a nondissociating compound whose empirical formula is CaH₂N, if 3.84 g of the compound in 500 g benzene give a freezing point depression of 0.307 °C? (The molal freezing point depression constant for benzene is 5.12 °C kg/mol.) ΔT = K_f * molality 0.307 °C = 5.12 °C kg/mol * molality molality = 0.0600 mol/kg molality = moles of solute / kg of solvent 0.0600 mol/kg = moles of solute / 0.500 kg solvent moles of solute = 0.0300 mol molar mass = 3.84 g solute / 0.0300 mol solute = 128 g/mol mass of molecular formula = 2 * mass of empirical formula = 2 * 64.08 g/mol = 128 g/mol Therefore, the molecular formula is C₈H₄N₂. 5) Liquids A and B form an ideal solution. The vapor pressure of pure A is 0.700 atm at the normal boiling point of a solution prepared from 0.250 mole of B and 0.650 mole of A. What is the vapor pressure of pure B at this temperature? Answer: 1.77 atm Ptotal = XA * P°A + XB * P°B 1 atm = (0.650 / 0.900) * (0.700 atm) + (0.250 / 0.900) * P°B 1 atm = 0.506 + 0.278 * P°B 0.494 = 0.278 * P°B P°B = 1.77 atm 6) A 0.900 L aqueous solution contains 30.0 g of a protein. The osmotic pressure of the solution is 12.7 torr at 25 °C. What is the molar mass of the protein? Answer: 4.88 x 10^4 g/mole π = MRT 12.7 torr (1 atm / 760 torr) = M * (0.0821 L atm / mol K) * 298 K M = 6.83 x 10^-4 mol/L no. of moles of protein = M * V = 6.83 x 10^-4 mol/L * 0.900 L = 6.15 x 10^-4 mol molar mass = 30.0 g protein / 6.15 x 10^-4 mol protein = 4.88 x 10^4 g/mol