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1 exercises12 questions solved

Exercise 1.1Types of Solutions, Concentration Terms and Vapour Pressure

Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Solution

Mass of benzene = 22 g, Mass of CCl₄ = 122 g Total mass of solution = 22 + 122 = 144 g Mass % of benzene = (22/144) × 100 = 15.28% Mass % of CCl₄ = (122/144) × 100 = 84.72% Check: 15.28 + 84.72 = 100% ✓

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Solution

Assume 100 g of solution: benzene = 30 g, CCl₄ = 70 g Molar mass of benzene (C₆H₆) = 6×12 + 6×1 = 78 g/mol Molar mass of CCl₄ = 12 + 4×35.5 = 154 g/mol Moles of benzene = 30/78 = 0.3846 mol Moles of CCl₄ = 70/154 = 0.4545 mol Total moles = 0.3846 + 0.4545 = 0.8391 Mole fraction of benzene = 0.3846/0.8391 = 0.459

Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L of solution; (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Solution

(a) M(Co(NO₃)₂·6H₂O) = 59 + 2(14+48) + 6×18 = 59 + 124 + 108 = 291 g/mol Moles = 30/291 = 0.1031 mol Molarity = 0.1031/4.3 = 0.024 M (b) Using M₁V₁ = M₂V₂: 0.5 × 30 = M₂ × 500 M₂ = 15/500 = 0.03 M

Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.

Solution

Molality m = 0.25 mol/kg Molar mass of urea = 60 g/mol For every 1 kg of water: moles of urea = 0.25 Mass of urea = 0.25 × 60 = 15 g Total solution mass = 1000 + 15 = 1015 g For 2.5 kg = 2500 g of solution: Mass of urea = (15/1015) × 2500 = 36.95 g ≈ 37 g

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/mL.

Solution

Molar mass of KI = 39 + 127 = 166 g/mol In 100 g solution: KI = 20 g, water = 80 g (a) Molality = moles of KI / kg of water = (20/166) / (80/1000) = 0.1205/0.080 = 1.506 m ≈ 1.51 m (b) Volume of 100 g solution = 100/1.202 = 83.19 mL = 0.08319 L Molarity = 0.1205/0.08319 = 1.449 M ≈ 1.45 M (c) Moles of water = 80/18 = 4.444 mol Moles of KI = 0.1205 mol Mole fraction of KI = 0.1205/(0.1205 + 4.444) = 0.1205/4.565 = 0.0264

H₂S, a toxic gas with rotten egg smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's law constant.

Solution

Solubility = 0.195 m = 0.195 mol/kg water Partial pressure of H₂S at STP = 1 bar = 0.987 atm ≈ 1 bar Moles of H₂S in 1000 g water = 0.195 mol Moles of water = 1000/18 = 55.56 mol Mole fraction of H₂S: χ(H₂S) = 0.195/(0.195 + 55.56) = 0.195/55.755 = 3.499 × 10⁻³ Henry's law: p = K_H × χ K_H = p/χ = 1 bar / (3.499 × 10⁻³) = 285.8 bar ≈ 286 bar

Henry's law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Solution

K_H = 1.67 × 10⁸ Pa, P = 2.5 atm = 2.5 × 101325 Pa = 2.533 × 10⁵ Pa Mole fraction of CO₂: χ(CO₂) = P/K_H = 2.533 × 10⁵/1.67 × 10⁸ = 1.517 × 10⁻³ Moles of water in 500 mL ≈ 500 g/18 g/mol = 27.78 mol For dilute solution: χ(CO₂) ≈ n(CO₂)/n(water) n(CO₂) = χ(CO₂) × n(water) = 1.517 × 10⁻³ × 27.78 = 0.04214 mol Mass of CO₂ = 0.04214 × 44 = 1.854 g ≈ 1.85 g

The vapour pressure of pure liquids A and B are 450 and 700 mmHg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mmHg. Also find the composition of the vapour phase.

Solution

P°_A = 450 mmHg, P°_B = 700 mmHg, P_total = 600 mmHg Using Raoult's law: P_total = P°_A × x_A + P°_B × x_B = P°_A × x_A + P°_B(1 – x_A) 600 = 450x_A + 700(1 – x_A) 600 = 450x_A + 700 – 700x_A –100 = –250x_A x_A = 0.4, x_B = 0.6 Partial pressures: P_A = 450 × 0.4 = 180 mmHg P_B = 700 × 0.6 = 420 mmHg Composition of vapour: y_A = P_A/P_total = 180/600 = 0.30 y_B = P_B/P_total = 420/600 = 0.70

Vapour pressure of pure water at 298 K is 23.8 mmHg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Solution

P° = 23.8 mmHg, mass of urea = 50 g, mass of water = 850 g M(urea) = 60 g/mol, M(water) = 18 g/mol Moles of urea = 50/60 = 0.833 mol Moles of water = 850/18 = 47.22 mol Mole fraction of urea: χ(urea) = 0.833/(0.833 + 47.22) = 0.833/48.053 = 0.01733 Relative lowering of vapour pressure: (P° – P)/P° = χ(solute) = 0.01733 Vapour pressure of solution: P = P°(1 – χ(urea)) = 23.8 × (1 – 0.01733) = 23.8 × 0.9827 = 23.39 mmHg

Q10

Boiling point of water at 750 mmHg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? Kb for water = 0.52 K kg mol⁻¹.

Solution

ΔTb = 100 – 99.63 = 0.37°C Kb = 0.52 K kg mol⁻¹, mass of water = 500 g = 0.5 kg M(sucrose, C₁₂H₂₂O₁₁) = 342 g/mol Using ΔTb = Kb × m: m = ΔTb/Kb = 0.37/0.52 = 0.7115 mol/kg Moles of sucrose = 0.7115 × 0.5 = 0.3558 mol Mass of sucrose = 0.3558 × 342 = 121.7 g ≈ 122 g

Q11

Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf for acetic acid = 3.9 K kg mol⁻¹.

Solution

ΔTf = 1.5°C, Kf = 3.9 K kg mol⁻¹, solvent mass = 75 g = 0.075 kg M(ascorbic acid) = 6×12 + 8×1 + 6×16 = 72+8+96 = 176 g/mol Using ΔTf = Kf × m: m = ΔTf/Kf = 1.5/3.9 = 0.3846 mol/kg Moles of ascorbic acid = 0.3846 × 0.075 = 0.02885 mol Mass = 0.02885 × 176 = 5.078 g ≈ 5.08 g

Q12

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185000 g mol⁻¹ in 450 mL of water at 37°C.

Solution

w = 1.0 g, M = 185000 g/mol, V = 450 mL = 0.45 L, T = 37°C = 310 K R = 8.314 J mol⁻¹ K⁻¹ Moles of polymer = 1.0/185000 = 5.405 × 10⁻⁶ mol Concentration C = 5.405 × 10⁻⁶/0.45 = 1.201 × 10⁻⁵ mol/L = 1.201 × 10⁻² mol/m³ Osmotic pressure: π = CRT = (5.405 × 10⁻⁶/0.00045) × 8.314 × 310 = 1.201 × 10⁻² × 8.314 × 310 = 30.93 Pa ≈ 30.9 Pa

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