A solution contains solute and solvent uniformly mixed. The solvent may be gas, liquid or solid and the solute may also be gas, liquid or solid. Examples include gas in gas: air; liquid in gas: moist air; solid in gas: camphor in air; gas in liquid: oxygen in water; liquid in liquid: ethanol in water; solid in liquid: sugar in water; gas in solid: hydrogen in palladium; liquid in solid: amalgam; solid in solid: brass.
A solution is a homogeneous mixture of two or more components. Depending on the physical states of solute and solvent, nine types are possible.
In this solid solution, palladium is the solid solvent and hydrogen gas is the solute absorbed in the metal lattice.
Hydrogen dissolved in palladium is an example.
For component i, mole fraction xi = ni/Σn. Molality m = moles of solute/kg of solvent. Molarity M = moles of solute/litre of solution. Mass percentage = (mass of component/mass of solution) x 100.
Mole fraction is moles of a component divided by total moles; molality is moles of solute per kg solvent; molarity is moles of solute per litre solution; mass percentage is mass of component per 100 mass units of solution.
Take 100 g solution. It contains 68 g HNO3 = 68/63 = 1.079 mol. Volume of 100 g solution = 100/1.504 = 66.49 mL = 0.06649 L. Molarity = 1.079/0.06649 = 16.23 M.
16.23 M.
In 100 g solution, glucose = 10 g and water = 90 g. Moles glucose = 10/180 = 0.0556 mol. Molality = 0.0556/0.090 = 0.617 m. Moles water = 90/18 = 5.00 mol, so xglucose = 0.0556/(0.0556 + 5.00) = 0.0110 and xwater = 0.989. Volume of 100 g solution = 100/1.2 = 83.33 mL = 0.08333 L. Molarity = 0.0556/0.08333 = 0.666 M.
Molality = 0.617 m; mole fraction of glucose = 0.0110 and of water = 0.989; molarity = 0.666 M.
For acid neutralisation, Na2CO3 + 2HCl → 2NaCl + H2O + CO2 and NaHCO3 + HCl → NaCl + H2O + CO2. Let the mixture contain x mol Na2CO3 and x mol NaHCO3. Its mass is 106x + 84x = 190x = 1 g, so x = 1/190 mol. Total HCl needed = 3x = 3/190 = 0.01579 mol. Volume of 0.1 M HCl = 0.01579/0.1 = 0.1579 L = 158 mL.
158 mL.
Solute in 300 g of 25% solution = 75 g. Solute in 400 g of 40% solution = 160 g. Total solute = 235 g and total solution = 700 g. Mass percentage = 235 x 100/700 = 33.6%.
33.6% by mass.
Molar mass of ethylene glycol = 62 g mol^-1. Moles = 222.6/62 = 3.59 mol. Mass of water = 0.200 kg, so molality = 3.59/0.200 = 17.9 m. Total mass = 422.6 g. Volume = 422.6/1.072 = 394.2 mL = 0.3942 L. Molarity = 3.59/0.3942 = 9.10 M.
Molality = 17.9 m; molarity = 9.10 M.
15 ppm by mass means 15 parts per 10^6 parts solution. Percent by mass = 15/10^6 x 100 = 0.0015%. In 1 kg water sample, mass of CHCl3 is approximately 15 mg = 0.015 g. Moles CHCl3 = 0.015/119.5 = 1.26 x 10^-4 mol. Molality ≈ 1.26 x 10^-4 m.
(i) 0.0015% by mass; (ii) 1.26 x 10^-4 m.
Alcohol and water molecules can form hydrogen bonds with one another. These solute-solvent interactions make lower alcohols highly soluble in water. However, the interactions differ from alcohol-alcohol and water-water interactions, so mixtures may show deviations from ideality.
Hydrogen bonding between alcohol and water controls miscibility and causes non-ideal behaviour.
For most gases, dissolving in a liquid releases heat. By Le Chatelier's principle, raising temperature favours the reverse process, namely escape of gas from solution. Therefore gas solubility decreases with increase in temperature.
Dissolution of gases in liquids is generally exothermic, so higher temperature lowers solubility.
Henry's law may be written as p = KH x, where p is partial pressure of the gas and x is its mole fraction in solution. Applications include carbonation of soft drinks under high CO2 pressure, controlling oxygen solubility in aquatic systems, explaining scuba-diving problems such as bends, and using helium-oxygen mixtures for deep-sea diving to reduce nitrogen absorption.
Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure over the solution at a given temperature.
At constant temperature, Henry's law gives p proportional to amount dissolved for the same solvent. Therefore p2/p1 = mass2/mass1 = (5.00 x 10^-2)/(6.56 x 10^-3). Thus p2 = 1 x 7.62 = 7.62 bar.
7.62 bar.
For a dilute solution with non-volatile solute, Raoult's law gives (p1° - p1)/p1° = x2, where x2 is the mole fraction of solute.
It is equal to the mole fraction of the solute.
For an ideal solution, ptotal = xA pA° + xB pB° = xA(450) + (1 - xA)(700). Since ptotal = 600, xA = 0.40 and xB = 0.60. Partial pressures are pA = 0.40 x 450 = 180 mm Hg and pB = 0.60 x 700 = 420 mm Hg. Vapour mole fractions are yA = 180/600 = 0.30 and yB = 420/600 = 0.70.
Liquid phase: xA = 0.40, xB = 0.60. Vapour phase: yA = 0.30, yB = 0.70.
Moles heptane = 26.0/100.2 = 0.259 mol and moles octane = 35.0/114.2 = 0.306 mol. Mole fraction heptane = 0.259/(0.259 + 0.306) = 0.459; octane = 0.541. Total vapour pressure = 0.459(105.2) + 0.541(46.8) = 73.6 kPa.
73.6 kPa.
A 1 molal aqueous solution contains 1 mol solute in 1 kg water. Moles water = 1000/18 = 55.5 mol. Mole fraction of water = 55.5/(55.5 + 1) = 0.9823. Vapour pressure = xwater p° = 0.9823 x 12.3 = 12.08 kPa.
12.08 kPa.
If vapour pressure is reduced to 80% of pure octane, xoctane = 0.80 and xsolute = 0.20. Moles octane = 114/114 = 1.00 mol. Let moles solute be n. Then 1/(1 + n) = 0.80, so n = 0.25 mol. Mass solute = 0.25 x 40 = 10.0 g.
10.0 g approximately.
Let moles of solute be n. Initially moles water = 90/18 = 5. After adding water, moles water = 108/18 = 6. Since p = xwater p°, the same p° applies: 2.8(n + 5)/5 = 2.9(n + 6)/6. Solving gives n = 30/23 mol. Molar mass = 30/n = 23 g mol^-1.
23 g mol^-1.
For equal mass percentages in water, freezing-point depression is proportional to moles of solute and therefore inversely proportional to molar mass. Cane sugar has molar mass 342 g mol^-1 and glucose has molar mass 180 g mol^-1. For cane sugar, ΔTf = 273.15 - 271 = 2.15 K. Hence ΔTf(glucose) = 2.15 x 342/180 = 4.09 K. Freezing point of glucose solution = 273.15 - 4.09 = 269.06 K.
269.06 K.
The formula masses follow AB2 → A + 2B and AB4 → A + 4B. For AB2, molar mass = Kf w x 1000/(ΔTf W) = 5.1 x 1 x 1000/(2.3 x 20) = 110.9 g mol^-1. For AB4, molar mass = 5.1 x 1 x 1000/(1.3 x 20) = 196.2 g mol^-1. Let atomic masses be A and B. A + 2B = 110.9 and A + 4B = 196.2. Hence 2B = 85.3, B = 42.6 and A = 25.6.
Atomic mass of A = 25.6 u and B = 42.6 u.
For dilute solutions at the same temperature, π = CRT. Therefore C = π/(RT) = 1.52/(0.08314 x 300) = 0.0609 mol L^-1. The given glucose data also confirms RT because 36 g glucose = 0.20 mol L^-1 and 4.98/0.20 ≈ 24.9.
0.061 mol L^-1.
Non-polar molecules such as hydrocarbons and I2/CCl4 interact mainly through dispersion forces. Ionic sodium perchlorate interacts with polar water through ion-dipole forces. Methanol can donate hydrogen bonds to acetone oxygen. Acetonitrile and acetone are polar molecules, so dipole-dipole interactions dominate.
(i) London dispersion; (ii) London dispersion; (iii) ion-dipole; (iv) hydrogen bonding/dipole-dipole; (v) dipole-dipole.
n-Octane is non-polar, so non-polar solutes dissolve best. KCl is ionic and least soluble. Methanol is strongly polar and hydrogen-bonded, so it is poorly soluble in octane. Acetonitrile is polar but less hydrogen-bonded than methanol. Cyclohexane is non-polar like n-octane and is most soluble.
KCl < CH3OH < CH3CN < cyclohexane.
Water dissolves compounds that can form strong hydrogen bonds or ion-like polar interactions. Formic acid and ethylene glycol are highly soluble because they form extensive hydrogen bonding. Phenol and pentanol have polar -OH groups but also sizeable hydrophobic parts, so they are only partially soluble. Toluene and chloroform are largely non-polar and are treated as insoluble in water.
Highly soluble: formic acid and ethylene glycol. Partially soluble: phenol and pentanol. Insoluble: toluene and chloroform.
Molality depends on moles of solute per kg solvent. Moles of Na+ = 92/23 = 4.0 mol in 1 kg water. Therefore molality = 4.0 m. The density is not required for molality.
4.0 mol kg^-1.
For CuS(s) ⇌ Cu2+ + S2-, if molar solubility is S, Ksp = [Cu2+][S2-] = S^2. Thus S = sqrt(6 x 10^-16) = 2.45 x 10^-8 M.
2.45 x 10^-8 M.
Mass percentage = mass of aspirin x 100/total mass of solution = 6.5 x 100/(6.5 + 450) = 1.42%.
1.42% by mass.
Molar mass of nalorphene = 311 g mol^-1. Moles in 1.5 mg = 1.5 x 10^-3/311 = 4.82 x 10^-6 mol. A 1.5 x 10^-3 m solution contains 1.5 x 10^-3 mol per kg water. Mass of water needed = (4.82 x 10^-6)/(1.5 x 10^-3) = 3.21 x 10^-3 kg = 3.21 g. The solute mass is tiny, so solution mass is about 3.21 g.
About 3.21 g of solution.
Moles required = M x V = 0.15 x 0.250 = 0.0375 mol. Molar mass of benzoic acid = 122 g mol^-1. Mass = 0.0375 x 122 = 4.58 g.
4.58 g.
Freezing-point depression is a colligative property and depends on the number of solute particles. Electron-withdrawing groups increase acid strength. Trifluoroacetic acid is strongest, then trichloroacetic acid, while acetic acid is weakest. Stronger acid dissociation gives a larger van't Hoff factor and hence a greater depression in freezing point.
The acids dissociate to different extents; stronger acids produce more ions and larger depression in freezing point.
For the weak acid dissociation, HA ⇌ H+ + A-. Molar mass of CH3CH2CHClCOOH = 122.5 g mol^-1. Molality = (10/122.5)/0.250 = 0.326 m. Ka = Cα^2/(1 - α) with C ≈ 0.326, giving α = 0.0634. Thus i = 1 + α = 1.063. Depression = iKf m = (1.063)(1.86)(0.326) = 0.646 K.
0.646 K.
Molar mass of CH2FCOOH = 78 g mol^-1. Moles = 19.5/78 = 0.250 mol; molality = 0.250/0.500 = 0.500 m. For water, ΔTf = iKf m, so i = 1.0/(1.86 x 0.500) = 1.08. For dissociation HA ⇌ H+ + A-, i = 1 + α, so α = 0.076. Ka = Cα^2/(1 - α) ≈ (0.500)(0.076)^2/(0.924) = 3.1 x 10^-3.
i = 1.08; Ka = 3.1 x 10^-3 approximately.
Moles glucose = 25/180 = 0.139 mol. Moles water = 450/18 = 25.0 mol. Mole fraction of water = 25.0/(25.0 + 0.139) = 0.9945. Vapour pressure = 0.9945 x 17.535 = 17.44 mm Hg.
17.44 mm Hg.
Henry's law gives mole fraction of oxygen, xO2 = p/KH = 0.4/(4.34 x 10^4) = 9.22 x 10^-6. One litre of water contains about 55.5 mol H2O, so dissolved oxygen concentration ≈ xO2 x 55.5 = 5.1 x 10^-4 mol L^-1.
5.1 x 10^-4 mol L^-1.
Moles A = 100/140 = 0.714; moles B = 1000/180 = 5.556. Hence xA = 0.1139 and xB = 0.8861. Partial pressure of B = xB pB° = 0.8861 x 500 = 443.0 torr. Therefore pA in solution = 475 - 443.0 = 32.0 torr. Since pA = xA pA°, pA° = 32.0/0.1139 = 280.6 torr.
pA° = 280.6 torr; pA = 32.0 torr.
Moles benzene = 80/78.11 = 1.024; moles toluene = 100/92.14 = 1.085. xbenzene = 1.024/(1.024 + 1.085) = 0.486. Partial pressures: pbenzene = 0.486 x 50.71 = 24.62 mm Hg; ptoluene = 0.514 x 32.06 = 16.49 mm Hg. Therefore ybenzene = 24.62/(24.62 + 16.49) = 0.599.
Mole fraction of benzene in vapour phase = 0.599.
Partial pressure equals mole fraction times total pressure. The ratio 78:21:1 gives mole fractions 0.78, 0.21 and 0.01. At total pressure 1 atm, the partial pressures are therefore 0.78 atm, 0.21 atm and 0.01 atm.
pN2 = 0.78 atm, pO2 = 0.21 atm and pAr = 0.01 atm.
Use πV = inRT. Moles CaCl2 = πV/(iRT) = (0.75 x 2.5)/(2.47 x 0.0821 x 300) = 0.0308 mol. Molar mass CaCl2 = 111 g mol^-1, so mass = 0.0308 x 111 = 3.42 g.
3.42 g CaCl2.
K2SO4 → 2K+ + SO4^2-, so complete dissociation gives i = 3. Moles K2SO4 = 0.025/174.3 = 1.43 x 10^-4 mol. Volume = 2 L, T = 298 K. π = i nRT/V = 3(1.43 x 10^-4)(0.0821)(298)/2 = 5.26 x 10^-3 atm.
5.26 x 10^-3 atm.