Intext Questions
1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Answer
Mass of solution = 22 + 122 = 144 g
Mass % of benzene = (22/144) × 100 = 15.28%
Mass % of CCl₄ = (122/144) × 100 = 84.72% (or 100 − 15.28)
2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Answer
Take 100 g solution: benzene = 30 g, CCl₄ = 70 g.
Moles of C₆H₆ = 30/78 = 0.385 mol; moles of CCl₄ = 70/154 = 0.455 mol
x(benzene) = 0.385/(0.385 + 0.455) = 0.385/0.840 = 0.459
⇒ x(CCl₄) = 1 − 0.459 = 0.541
3. 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.
Answer
(a) Molar mass of Co(NO₃)₂·6H₂O = 59 + 2(14 + 48) + 6(18) = 291 g mol⁻¹
⇒ moles = 30/291 = 0.103 mol
⇒ Molarity = 0.103/4.3 = 0.024 M
(b) Using M₁V₁ = M₂V₂
⇒ 0.5 × 30 = M₂ × 500
⇒ M₂ = 0.03 M
4. Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Answer
0.25 molal means 0.25 mol urea in 1000 g water. Molar mass of urea = 60 g mol⁻¹
⇒ mass of urea = 0.25 × 60 = 15 g
Mass of this solution = 1000 + 15 = 1015 g
Urea needed for 2500 g solution = (15/1015) × 2500 = 36.95 g
5. 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⁻¹.
Answer
Take 100 g solution: KI = 20 g, water = 80 g. Molar mass KI = 166 g mol⁻¹
⇒ moles KI = 20/166 = 0.1205 mol
(a) Molality = 0.1205/0.080 kg = 1.506 mol kg⁻¹
(b) Volume of solution = 100/1.202 = 83.2 mL = 0.0832 L
⇒ Molarity = 0.1205/0.0832 = 1.45 M
(c) Moles of water = 80/18 = 4.444
⇒ x(KI) = 0.1205/(0.1205 + 4.444) = 0.0263
6. H₂S, a toxic gas with rotten egg like 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.
Answer
0.195 m means 0.195 mol H₂S in 1 kg (55.56 mol) water.
x(H₂S) = 0.195/(0.195 + 55.56) = 0.0035
At STP, p = 0.987 bar. Henry's law: p = K_H x
⇒ K_H = 0.987/0.0035 = 282 bar
7. 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.
Answer
p(CO₂) = 2.5 atm = 2.5 × 1.01325 × 10⁵ Pa = 2.533 × 10⁵ Pa
x(CO₂) = p/K_H = 2.533 × 10⁵ / 1.67 × 10⁸ = 1.517 × 10⁻³
500 mL water ≈ 500 g
⇒ n(H₂O) = 500/18 = 27.78 mol. Since x is very small, n(CO₂) ≈ x × n(H₂O) = 1.517 × 10⁻³ × 27.78 = 0.0421 mol
Mass of CO₂ = 0.0421 × 44 = 1.85 g
8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Answer
p(total) = p°A xA + p°B (1 − xA)
⇒ 600 = 450 xA + 700(1 − xA)
⇒ 250 xA = 100
⇒ xA = 0.40, xB = 0.60
pA = 450 × 0.4 = 180 mm Hg; pB = 700 × 0.6 = 420 mm Hg
Vapour phase: yA = 180/600 = 0.30; yB = 420/600 = 0.70
9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 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.
Answer
n(urea) = 50/60 = 0.833 mol; n(water) = 850/18 = 47.22 mol
Relative lowering = (p° − p)/p° = x(urea) = 0.833/(0.833 + 47.22) = 0.0173
p = p°(1 − 0.0173) = 23.8 × 0.9827 = 23.4 mm Hg
10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Answer
ΔTb = 100 − 99.63 = 0.37 K; Kb (water) = 0.52 K kg mol⁻¹; M (sucrose) = 342 g mol⁻¹
w₂ = ΔTb × M₂ × w₁ / (Kb × 1000) = (0.37 × 342 × 500)/(0.52 × 1000) = 121.67 g
11. 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 = 3.9 K kg mol⁻¹.
Answer
M (C₆H₈O₆) = 176 g mol⁻¹
w₂ = ΔTf × M₂ × w₁ / (Kf × 1000) = (1.5 × 176 × 75)/(3.9 × 1000) = 5.08 g
12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Answer
n = 1.0/185000 = 5.41 × 10⁻⁶ mol; V = 0.450 L; T = 310 K; R = 8.314 kPa L K⁻¹ mol⁻¹
π = nRT/V = (5.41 × 10⁻⁶ × 8.314 × 310)/0.450 kPa = 0.03096 kPa = 30.96 Pa
Exercises
1. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer
A solution is a homogeneous mixture of two or more components whose composition is uniform throughout. The component present in the larger amount is the solvent, the others are solutes. In a binary solution, the solvent can be a gas, liquid or solid, giving three main types (nine combinations):
(i) Gaseous solutions (solvent gas): gas in gas – mixture of O₂ and N₂; liquid in gas – chloroform mixed with N₂ gas; solid in gas – camphor in N₂ gas.
(ii) Liquid solutions (solvent liquid): gas in liquid – O₂ dissolved in water; liquid in liquid – ethanol in water; solid in liquid – glucose in water.
(iii) Solid solutions (solvent solid): gas in solid – H₂ in palladium; liquid in solid – amalgam of mercury with sodium; solid in solid – copper dissolved in gold.
2. Give an example of a solid solution in which the solute is a gas.
Answer
Solution of hydrogen in palladium (hydrogen gas is absorbed by palladium metal).
3. Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
Answer
(i) Mole fraction (x) of a component = moles of that component ÷ total moles of all components. For a binary solution x₁ + x₂ = 1.
(ii) Molality (m) = moles of solute ÷ mass of solvent in kg (unit mol kg⁻¹); it does not change with temperature.
(iii) Molarity (M) = moles of solute ÷ volume of solution in litres (unit mol L⁻¹); it changes with temperature because volume changes.
(iv) Mass percentage (w/w) of a component = (mass of component ÷ total mass of solution) × 100.
4. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹?
Answer
100 g solution has 68 g HNO₃
⇒ moles = 68/63 = 1.079 mol
Volume = 100/1.504 = 66.49 mL = 0.06649 L
Molarity = 1.079/0.06649 = 16.23 M
5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL⁻¹, then what shall be the molarity of the solution?
Answer
100 g solution: glucose 10 g (10/180 = 0.0556 mol), water 90 g (90/18 = 5 mol)
Molality = 0.0556/0.090 = 0.617 mol kg⁻¹
x(glucose) = 0.0556/5.0556 = 0.011; x(water) = 1 − 0.011 = 0.989
Volume = 100/1.2 = 83.33 mL
⇒ Molarity = 0.0556/0.08333 = 0.67 M
6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?
Answer
Let moles of each = x. 106x + 84x = 1
⇒ x = 1/190 = 5.263 × 10⁻³ mol
Na₂CO₃ + 2HCl
⇒ 2NaCl + H₂O + CO₂ (needs 2x); NaHCO₃ + HCl
⇒ NaCl + H₂O + CO₂ (needs x)
Total HCl = 3x = 0.01579 mol
⇒ Volume = 0.01579/0.1 = 0.1579 L = 157.9 mL ≈ 158 mL
7. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer
Solute = 0.25 × 300 + 0.40 × 400 = 75 + 160 = 235 g; total mass = 700 g
Mass % of solute = (235/700) × 100 = 33.57%; mass % of solvent = 66.43%
8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL⁻¹, then what shall be the molarity of the solution?
Answer
M(C₂H₆O₂) = 62 g mol⁻¹
⇒ moles = 222.6/62 = 3.59 mol
Molality = 3.59/0.200 = 17.95 mol kg⁻¹
Mass of solution = 422.6 g
⇒ volume = 422.6/1.072 = 394.2 mL
⇒ Molarity = 3.59/0.3942 = 9.11 M
9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample.
Answer
15 ppm = 15 g CHCl₃ in 10⁶ g solution.
(i) Mass % = (15/10⁶) × 100 = 1.5 × 10⁻³ %
(ii) Moles CHCl₃ = 15/119.5 = 0.1255 mol; mass of water ≈ 10⁶ g = 1000 kg
⇒ molality = 0.1255/1000 = 1.25 × 10⁻⁴ mol kg⁻¹
10. What role does the molecular interaction play in a solution of alcohol and water?
Answer
Both alcohol and water are strongly hydrogen bonded in the pure state. On mixing, the new alcohol–water hydrogen bonds are weaker than the alcohol–alcohol and water–water interactions they replace. Molecules therefore escape more easily, so the solution shows a positive deviation from Raoult's law: its vapour pressure is higher (and boiling point lower) than expected, mixing is slightly endothermic and volume increases slightly.
11. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer
Dissolution of a gas in a liquid is an exothermic process (gas molecules lose freedom and energy when they enter the liquid). By Le Chatelier's principle, raising the temperature shifts the equilibrium gas + liquid ⇌ solution backwards, so solubility falls. Equivalently, the Henry's law constant K_H increases with temperature, so at a given pressure the mole fraction of dissolved gas decreases.
12. State Henry's law and mention some important applications.
Answer
Henry's law: at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid; equivalently, the partial pressure of the gas in the vapour phase is proportional to its mole fraction in solution: p = K_H x.
Applications: (i) Soft drinks and soda water are sealed under high CO₂ pressure to increase its solubility. (ii) Scuba divers breathing compressed air have more N₂ dissolved in blood at depth; on rapid ascent it forms bubbles causing "bends", so air tanks are diluted with helium (11.7% He, 56.2% N₂, 32.1% O₂). (iii) At high altitudes low partial pressure of O₂ lowers oxygen in blood, causing anoxia (weakness, inability to think clearly).
13. The partial pressure of ethane over a solution containing 6.56 × 10⁻³ g of ethane is 1 bar. If the solution contains 5.00 × 10⁻² g of ethane, then what shall be the partial pressure of the gas?
Answer
By Henry's law, mass dissolved ∝ partial pressure: m₁/m₂ = p₁/p₂
p₂ = p₁ × m₂/m₁ = 1 bar × (5.00 × 10⁻²)/(6.56 × 10⁻³) = 7.62 bar
14. What is meant by positive and negative deviations from Raoult's law and how is the sign of ΔmixH related to positive and negative deviations from Raoult's law?
Answer
Positive deviation: the observed vapour pressure of each component (and the total) is higher than predicted by Raoult's law. It happens when A–B interactions are weaker than A–A and B–B interactions, so molecules escape more easily; e.g. ethanol + acetone, CS₂ + acetone. Here ΔmixH is positive (endothermic) and ΔmixV is positive.
Negative deviation: the observed vapour pressure is lower than predicted, because A–B interactions are stronger than A–A and B–B; e.g. chloroform + acetone (H-bonding), phenol + aniline, HNO₃ + water. Here ΔmixH is negative (exothermic) and ΔmixV is negative.
15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer
At the normal boiling point p°(water) = 1.013 bar; p = 1.004 bar. 100 g solution: w₂ = 2 g, w₁ = 98 g.
(p° − p)/p° = (w₂ × M₁)/(M₂ × w₁) (dilute solution)
⇒ (1.013 − 1.004)/1.013 = (2 × 18)/(M₂ × 98)
0.00888 = 36/(98 M₂)
⇒ M₂ = 41.35 g mol⁻¹
16. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer
Heptane C₇H₁₆ = 100 g mol⁻¹
⇒ 26/100 = 0.26 mol; octane C₈H₁₈ = 114 g mol⁻¹
⇒ 35/114 = 0.307 mol
x(heptane) = 0.26/0.567 = 0.459; x(octane) = 0.541
p = 105.2 × 0.459 + 46.8 × 0.541 = 48.3 + 25.3 = 73.6 kPa
17. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer
1 molal: 1 mol solute in 1000 g (55.56 mol) water
⇒ x(solute) = 1/56.56 = 0.0177
p = p° × x(water) = 12.3 × (1 − 0.0177) = 12.08 kPa
18. Calculate the mass of a non-volatile solute (molar mass 40 g mol⁻¹) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer
p = 0.80 p°
⇒ relative lowering = 0.20 = x(solute). Moles of octane = 114/114 = 1 mol.
n₂/(n₂ + 1) = 0.20
⇒ n₂ = 0.25 mol
⇒ mass = 0.25 × 40 = 10 g
(If the dilute-solution approximation n₂/n₁ = 0.20 is used, n₂ = 0.20 mol and mass = 8 g.)
19. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Answer
Let molar mass = M, n₂ = 30/M. Using p/p° = n₁/(n₁ + n₂):
Case 1: n₁ = 90/18 = 5
⇒ 2.8/p° = 5/(5 + 30/M)
Case 2: n₁ = 108/18 = 6
⇒ 2.9/p° = 6/(6 + 30/M)
Dividing: 2.8/2.9 = 5(6 + 30/M)/[6(5 + 30/M)]
⇒ 16.8(5 + 30/M) = 14.5(6 + 30/M)
⇒ 2.3 × 30/M = 3
⇒ 30/M = 1.304
(i) M = 23 g mol⁻¹
(ii) p° = 2.8 × (5 + 1.304)/5 = 3.53 kPa
20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Answer
Sugar: ΔTf = 273.15 − 271 = 2.15 K; molality = (5/342)/0.095 = 0.1539 m
⇒ Kf = 2.15/0.1539 = 13.97 K kg mol⁻¹ (apparent value from data)
Glucose: molality = (5/180)/0.095 = 0.2924 m
⇒ ΔTf = 13.97 × 0.2924 = 4.09 K
Freezing point = 273.15 − 4.09 = 269.06 K
(Shortcut: ΔTf ∝ 1/M for equal mass %, so ΔTf(glucose) = 2.15 × 342/180 = 4.09 K.)
21. Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol⁻¹. Calculate atomic masses of A and B.
Answer
M₂ = Kf × w₂ × 1000/(ΔTf × w₁)
M(AB₂) = (5.1 × 1 × 1000)/(2.3 × 20) = 110.87 g mol⁻¹
M(AB₄) = (5.1 × 1 × 1000)/(1.3 × 20) = 196.15 g mol⁻¹
a + 2b = 110.87 and a + 4b = 196.15
⇒ 2b = 85.28
⇒ b = 42.64; a = 110.87 − 85.28 = 25.59
Atomic mass of A ≈ 25.59 u, of B ≈ 42.64 u.
22. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer
π = CRT, so at the same T, π₁/π₂ = C₁/C₂. C₁ = 36/180 = 0.2 mol L⁻¹
C₂ = 0.2 × 1.52/4.98 = 0.061 mol L⁻¹
23. Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I₂ and CCl₄ (iii) NaClO₄ and water (iv) methanol and acetone (v) acetonitrile (CH₃CN) and acetone (C₃H₆O).
Answer
(i) London dispersion forces (both non-polar). (ii) London dispersion forces (both non-polar). (iii) Ion–dipole interaction. (iv) Hydrogen bonding (O–H of methanol with C=O oxygen of acetone), along with dipole–dipole. (v) Dipole–dipole interaction (both polar, no H-bond donor).
24. Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer
n-Octane is non-polar, so "like dissolves like": the more non-polar the solute, the higher its solubility. KCl is ionic (least soluble), CH₃OH is polar and strongly H-bonded, CH₃CN is polar but less so in terms of self-association, and cyclohexane is non-polar.
Order: KCl < CH₃OH < CH₃CN < Cyclohexane
25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer
Insoluble: toluene, chloroform (non-polar / weakly polar, no H-bonding with water).
Partially soluble: phenol, pentanol (an –OH group forms H-bonds, but the large hydrocarbon part is hydrophobic).
Highly soluble: formic acid, ethylene glycol (small molecules that form extensive H-bonds with water).
26. If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molarity of Na⁺ ions in the lake.
Answer
Moles of Na⁺ = 92/23 = 4 mol per kg of lake water.
Volume of 1 kg lake water = 1000/1.25 = 800 mL = 0.8 L
⇒ Molarity = 4/0.8 = 5 M
(If the 92 g is added to 1 kg of pure water, solution mass = 1092 g, V = 873.6 mL and M ≈ 4.58 M; the textbook-style answer is 5 M.)
27. If the solubility product of CuS is 6 × 10⁻¹⁶, calculate the maximum molarity of CuS in aqueous solution.
Answer
CuS ⇌ Cu²⁺ + S²⁻; Ksp = s × s = s²
s = √(6 × 10⁻¹⁶) = 2.45 × 10⁻⁸ mol L⁻¹ (maximum molarity)
28. Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Answer
Mass of solution = 6.5 + 450 = 456.5 g
⇒ mass % of aspirin = (6.5/456.5) × 100 = 1.42%
29. Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10⁻³ m aqueous solution required for the above dose.
Answer
M(C₁₉H₂₁NO₃) = 19(12) + 21(1) + 14 + 3(16) = 311 g mol⁻¹
Moles in dose = 1.5 × 10⁻³ g/311 = 4.82 × 10⁻⁶ mol
Mass of water needed = 4.82 × 10⁻⁶/1.5 × 10⁻³ = 3.21 × 10⁻³ kg = 3.21 g
Mass of solution ≈ 3.21 g + 0.0015 g ≈ 3.21 g
30. Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer
Moles = 0.15 × 0.250 = 0.0375 mol; M(C₆H₅COOH) = 122 g mol⁻¹
⇒ mass = 0.0375 × 122 = 4.575 g
31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer
Depression in freezing point depends on the number of particles (van't Hoff factor). Cl and F are electron-withdrawing (–I effect), which stabilises the carboxylate ion and increases acid strength. F has a stronger –I effect than Cl. Hence acid strength and degree of ionisation increase as CH₃COOH < CCl₃COOH < CF₃COOH, so the number of ions (i) and therefore ΔTf increase in the same order.
32. Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. Ka = 1.4 × 10⁻³, Kf = 1.86 K kg mol⁻¹.
Answer
M(C₄H₇ClO₂) = 122.5 g mol⁻¹
⇒ moles = 10/122.5 = 0.0816 mol
⇒ m = 0.0816/0.250 = 0.3265 mol kg⁻¹
Ka = cα²/(1 − α)
⇒ 0.3265α² + 1.4 × 10⁻³ α − 1.4 × 10⁻³ = 0
⇒ α = 0.0634
i = 1 + α = 1.0634 (HA ⇌ H⁺ + A⁻)
ΔTf = i Kf m = 1.0634 × 1.86 × 0.3265 = 0.65 K
33. 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
Answer
M(CH₂FCOOH) = 78 g mol⁻¹
⇒ moles = 19.5/78 = 0.25 mol
⇒ m = 0.25/0.5 = 0.5 mol kg⁻¹
Calculated ΔTf = Kf m = 1.86 × 0.5 = 0.93 K
⇒ i = observed/calculated = 1.0/0.93 = 1.0753
For HA ⇌ H⁺ + A⁻, i = 1 + α
⇒ α = 0.0753
Ka = cα²/(1 − α) = 0.5 × (0.0753)²/0.9247 = 3.07 × 10⁻³
34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer
n(glucose) = 25/180 = 0.1389 mol; n(water) = 450/18 = 25 mol
⇒ x(glucose) = 0.1389/25.139 = 0.00553
p = 17.535 × (1 − 0.00553) = 17.44 mm Hg
35. Henry's law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer
p = K_H x
⇒ x = p/K_H = 760/(4.27 × 10⁵) = 1.78 × 10⁻³
Solubility of methane (mole fraction) = 1.78 × 10⁻³
36. 100 g of liquid A (molar mass 140 g mol⁻¹) was dissolved in 1000 g of liquid B (molar mass 180 g mol⁻¹). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer
nA = 100/140 = 0.714 mol; nB = 1000/180 = 5.556 mol
⇒ xA = 0.714/6.270 = 0.114; xB = 0.886
pB = 500 × 0.886 = 443 torr
⇒ pA = 475 − 443 = 32 torr (vapour pressure of A in solution)
p°A = pA/xA = 32/0.114 ≈ 280.7 torr (pure A)
37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p(total), p(chloroform), and p(acetone) as a function of x(acetone). The experimental data observed for different compositions of mixture is: 100 × x(acetone): 0, 11.8, 23.4, 36.0, 50.8, 58.2, 64.5, 72.1; p(acetone)/mm Hg: 0, 54.9, 110.1, 202.4, 322.7, 405.9, 454.1, 521.1; p(chloroform)/mm Hg: 632.8, 548.1, 469.4, 359.7, 257.7, 193.6, 161.2, 120.7. Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
Answer
Ideal values: p(acetone) = 741.8 x, p(chloroform) = 632.8(1 − x); these are straight lines from 0 to 741.8 and from 632.8 to 0, and ideal p(total) is a straight line from 632.8 (x = 0) to 741.8 (x = 1).
Experimental p(total) = p(acetone) + p(chloroform): at x = 0, 0.118, 0.234, 0.360, 0.508, 0.582, 0.645, 0.721
⇒ 632.8, 603.0, 579.5, 562.1, 580.4, 599.5, 615.3, 641.8 mm Hg.
Ideal p(total) at the same x
⇒ 632.8, 645.7, 658.3, 672.0, 688.2, 696.2, 703.1, 711.4 mm Hg.
The experimental total (and each partial pressure) lies below the ideal line and the total curve shows a minimum near x(acetone) ≈ 0.36. Hence the mixture shows negative deviation from Raoult's law, due to hydrogen bonding between the C–H of chloroform and the C=O oxygen of acetone.
38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer
n(benzene) = 80/78 = 1.026 mol; n(toluene) = 100/92 = 1.087 mol
⇒ x(benzene) = 1.026/2.113 = 0.486; x(toluene) = 0.514
p(benzene) = 50.71 × 0.486 = 24.62 mm Hg; p(toluene) = 32.06 × 0.514 = 16.49 mm Hg
⇒ p(total) = 41.11 mm Hg
y(benzene) = 24.62/41.11 = 0.60
39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Answer
Total pressure = 10 atm = 7600 mm Hg. Partial pressures: p(O₂) = 0.20 × 7600 = 1520 mm; p(N₂) = 0.79 × 7600 = 6004 mm
x(O₂) = 1520/(3.30 × 10⁷) = 4.61 × 10⁻⁵
x(N₂) = 6004/(6.51 × 10⁷) = 9.22 × 10⁻⁵
40. Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Answer
π = i(n/V)RT
⇒ n = πV/(iRT) = (0.75 × 2.5)/(2.47 × 0.0821 × 300) = 1.875/60.83 = 0.0308 mol
Mass of CaCl₂ = 0.0308 × 111 = 3.42 g
41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer
K₂SO₄
⇒ 2K⁺ + SO₄²⁻, so i = 3. M(K₂SO₄) = 174 g mol⁻¹
⇒ n = 0.025/174 = 1.437 × 10⁻⁴ mol
π = i(n/V)RT = 3 × (1.437 × 10⁻⁴/2) × 0.0821 × 298 = 5.27 × 10⁻³ atm
Important Extra Questions
Why is molality preferred over molarity in expressing concentration for colligative property measurements?
Answer
Molality is based on masses, which do not change with temperature, whereas molarity depends on volume, which changes with temperature. So molality gives temperature-independent values.
What is an azeotrope? Give one example each of minimum and maximum boiling azeotropes.
Answer
An azeotrope is a binary mixture that boils at constant temperature with the same composition in liquid and vapour, so it cannot be separated by fractional distillation. Minimum boiling: ethanol–water (about 95% ethanol by volume), shown by large positive deviation. Maximum boiling: nitric acid–water (about 68% HNO₃ by mass), shown by large negative deviation.
What are isotonic solutions? Why does a blood cell shrink in a hypertonic solution?
Answer
Isotonic solutions have the same osmotic pressure at a given temperature (e.g. 0.9% NaCl is isotonic with blood fluid). In a hypertonic solution (more concentrated outside), water flows out of the cell by osmosis, so the cell shrinks.
Why is osmotic pressure the preferred method for finding molar masses of proteins and polymers?
Answer
It is measured at room temperature, so biomolecules are not damaged by heating, and even very dilute solutions give a measurable pressure, whereas ΔTb or ΔTf would be too small to measure for such large molar masses.
What is the van't Hoff factor for K₄[Fe(CN)₆] assuming complete dissociation, and what will be its value for acetic acid in benzene?
Answer
K₄[Fe(CN)₆]
⇒ 4K⁺ + [Fe(CN)₆]⁴⁻ gives 5 ions, so i = 5. Acetic acid dimerises in benzene through hydrogen bonding, so i is less than 1 (about 0.5 for complete dimerisation).