Corrected exercises: solutions and colligative properties (202-SN1-RE)

$n=\dfrac{25.0}{58.44}=0.4278$ mol. Molarity relates this amount to the volume of SOLUTION: $C=\dfrac{0.4278}{0.250}=1.71$ mol/L.

The mass of the solution is $m=\rho V=1.05\times 250=262.5$ g. The solute accounts for 25.0 g of it, so the solvent accounts for $262.5-25.0=237.5$ g, or 0.2375 kg. Molality relates the amount to the mass of SOLVENT: $m=\dfrac{0.4278}{0.2375}=1.80$ mol/kg.

Mass percent: $\dfrac{25.0}{262.5}\times 100=9.52$ %. For the mole fraction, we need the amount of water: $n\left(H_{2}O\right)=\dfrac{237.5}{18.02}=13.18$ mol. Then $x(NaCl)=\dfrac{0.4278}{0.4278+13.18}=\dfrac{0.4278}{13.61}=0.0314$. The salt therefore makes up only 3.1% of the molecules, although it accounts for 9.5% of the mass: the two percentages have nothing to do with each other.

They differ because the denominator is not the same: 0.250 L of solution against 0.2375 kg of water. Because the solution is denser than pure water and the solute takes up part of the volume, we get $1.71$ against $1.80$ here, a 5% difference. The two values get closer when the solution is DILUTE and the solvent is water: at low concentration, the density tends to 1.00 g/mL and the mass of solvent in kilograms tends to the volume of solution in litres. In practice, below 0.1 mol/L, mixing them up has no numerical consequence, which is why the confusion goes unnoticed for so long.

The content is $\dfrac{5.0\text{ mg}}{2.0\text{ L}}=2.5$ mg/L, or 2.5 ppm, since in dilute aqueous solution 1 mg/L equals 1 ppm, a litre having a mass of approximately 1 kg. This value is 500 times the guideline of 0.005 mg/L: the water is far from fit to drink. The ppm unit is used precisely because the corresponding molarities are awkward to handle, here $1.2\times 10^{-5}$ mol/L.

The hydroxyl group is polar and carries a hydrogen bonded to an oxygen: it can donate and accept hydrogen bonds with water molecules, exactly as water does with itself. The carbon chain, by contrast, is nonpolar and can only form London dispersion forces; inserting it into water means breaking hydrogen bonds between water molecules while offering nothing in exchange. Each alcohol molecule is therefore the scene of a competition between a hydrophilic head and a hydrophobic tail. As long as the chain is short, the head wins and the miscibility is total; from four carbons on, the tail takes over and the solubility collapses. Between 1-butanol and 1-octanol it is divided by $\dfrac{7.9}{0.05}=158$, for only four more carbons.

The trend reverses. In hexane there are only London forces, and it is the carbon chain that fits in naturally; it is the hydroxyl group that becomes a nuisance, since it loses the hydrogen bonds it formed in the pure alcohol without being able to form new ones with the solvent. Long-chain alcohols such as octanol are therefore very soluble in it, and it is the shortest ones, methanol first, that are the least soluble. A single substance can thus be “very soluble” or “sparingly soluble” depending on the solvent: solubility is never a property of the solute alone.

Vitamin C, bristling with hydroxyl groups, is water-soluble: it circulates in the blood and the excess is quickly eliminated by the kidneys. Vitamin A, dominated by its long hydrocarbon chain, is fat-soluble: it accumulates in fatty tissue and in the liver. The practical consequence is clear. An excess of vitamin C is not very dangerous, since the body gets rid of it within a few hours, which also means it has to be consumed regularly. An excess of vitamin A, on the other hand, accumulates with no rapid route of elimination and becomes toxic: hypervitaminosis A is a real risk with dietary supplements, whereas hypervitaminosis C practically does not exist. The same reasoning applies to vitamins D, E and K, which are fat-soluble, and to the B vitamins, which are water-soluble.

The stearate ion is amphiphilic: its charged head is strongly hydrated, and its seventeen-carbon tail can only live in a nonpolar environment. In the presence of water and fat, these ions organize into micelles, spheres in which all the tails point inward and enclose the fat droplets, while all the charged heads form the outer surface in contact with water. From the outside, the micelle looks like one big ion: it is therefore perfectly dispersed in water, and the fat leaves with it in the rinse. Soap does not make fat soluble, it wraps it in a soluble envelope. As for diiodine, it is a nonpolar molecule whose molecules hold together only through London forces: cyclohexane, also nonpolar, replaces them identically, whereas water would have to break its network of hydrogen bonds with no compensation. Sodium chloride, for its part, is an ionic solid whose lattice can only be taken apart by a solvent capable of strongly hydrating the ions, which only a very polar molecule can do: water dissolves it, cyclohexane is totally incapable of doing so. It is the same rule applied twice in opposite directions.

The partial pressure of oxygen in air is $0.210\times 1.00=0.210$ atm, so $S=1.3\times 10^{-3}\times 0.210=2.73\times 10^{-4}$ mol/L. By mass: $2.73\times 10^{-4}\times 32.00=8.7\times 10^{-3}$ g/L, or 8.7 mg/L. The agreement with the 8 mg/L measured is excellent, which validates both Henry's law and the fact that oxygen does not react with water. Keep the order of magnitude in mind: all aquatic life depends on less than ten milligrams of oxygen per litre, roughly thirty times less than the same volume of air contains.

Before opening: $S=3.4\times 10^{-2}\times 4.0=0.136$ mol/L. Once the bottle is open, the gas above the liquid is ordinary air and the $CO_{2}$ partial pressure falls to $4.0\times 10^{-4}$ atm, so $S=3.4\times 10^{-2}\times 4.0\times 10^{-4}=1.4\times 10^{-5}$ mol/L, ten thousand times less. In the open air the drink is therefore heavily supersaturated, and it loses gas until it goes flat. The gas released per litre is $0.136-0.000014\approx 0.136$ mol, that is $V=\dfrac{nRT}{P}=\dfrac{0.136\times 8.314\times 298}{101.3}=3.3$ L of carbon dioxide: more than three times the volume of the bottle.

The drop is $\dfrac{10.1-8.3}{10.1}\times 100=17.8\%$ for only ten degrees. Dissolving a gas is exothermic, since free gas molecules end up trapped and solvated: by Le Chatelier's principle, raising the temperature shifts this equilibrium in the direction that absorbs heat, that is, toward outgassing. Thermal agitation adds to this, helping dissolved molecules leave the liquid. For a solid, on the contrary, dissolution is most often endothermic and heating helps it. Direct ecological consequence: a power plant that discharges its cooling water warms the river and removes nearly a fifth of its dissolved oxygen, at the very moment when the fish, whose metabolism speeds up with temperature, need more of it. This is thermal pollution, and it kills without any toxic substance having been released.

The partial pressure of nitrogen goes from $0.790\times 1.0=0.79$ atm at the surface to $0.790\times 4.0=3.16$ atm at 30 m. Since Henry's law is a simple proportionality, the amount of nitrogen dissolved in the blood and tissues is multiplied by 4.0. This nitrogen serves no purpose, it just dissolves. If the diver ascends slowly, the pressure falls gradually and the excess leaves through the lungs at the rate at which it is released. If the diver ascends quickly, the solution suddenly becomes supersaturated, exactly like the opened bottle of question b), and the nitrogen forms bubbles in the blood and joints: this is decompression sickness. Decompression stops have no other purpose than to give Henry equilibrium time to re-establish itself without bubble formation.

Dilution conserves the amount of substance: $C_{1}V_{1}=C_{2}V_{2}$, so $V_{1}=\dfrac{C_{2}V_{2}}{C_{1}}=\dfrac{0.150\times 0.500}{12.0}=6.25\times 10^{-3}$ L, or 6.25 mL. Procedure: draw 6.25 mL of the commercial solution with a pipette, transfer it to a 500 mL volumetric flask that ALREADY contains some distilled water, then fill up to the calibration mark and mix. Never put the acid into an empty flask that you then fill with water.

The reaction is $HCl+NaOH\rightarrow NaCl+H_{2}O$, in a 1 to 1 ratio. Then $n(HCl)=0.0187\times 0.200=3.74\times 10^{-3}$ mol, so $n(NaOH)=3.74\times 10^{-3}$ mol as well, and $C(NaOH)=\dfrac{3.74\times 10^{-3}}{0.0250}=0.150$ mol/L.

$Ag^{+}+Cl^{-}\rightarrow AgCl(s)$. Silver nitrate is the limiting reactant since $NaCl$ is in excess: $n\left(Ag^{+}\right)=0.0500\times 0.100=5.00\times 10^{-3}$ mol, so $n(AgCl)=5.00\times 10^{-3}$ mol and $m=5.00\times 10^{-3}\times 143.32=0.717$ g.

Because dissolving concentrated sulfuric or hydrochloric acid is strongly EXOTHERMIC. When water is added to the acid, the first drop of water receives all the heat released in a very small volume: it can instantly reach boiling and spatter concentrated acid. When the acid is instead added to a large volume of water, the heat is spread through the whole mass of water, whose high heat capacity limits the temperature rise. The mnemonic is “always add acid to water, never the reverse”.

$\Delta T_{f}=i\,K_{f}\,m$. Glucose, a nonelectrolyte, $i=1$: $\Delta T_{f}=1\times 1.86\times 1.00=1.86$ °C, so it freezes at $-1.86$ °C. $NaCl$, $i=2$: $\Delta T_{f}=3.72$ °C, so $-3.72$ °C. $CaCl_{2}$, $i=3$ because it releases one $Ca^{2+}$ and two $Cl^{-}$: $\Delta T_{f}=5.58$ °C, so $-5.58$ °C.

The ranking follows the van't Hoff factor exactly, and therefore the NUMBER OF PARTICLES in solution. At equal molality, a mole of glucose gives one mole of particles, a mole of $NaCl$ gives two, a mole of $CaCl_{2}$ gives three. That is the very meaning of the word “colligative”: the property is tied to counting, not to chemical identity. A chloride ion and a glucose molecule have rigorously the same effect on the freezing point.

We look for $m$ such that $2\times 1.86\times m=10.0$, so $m=\dfrac{10.0}{3.72}=2.69$ mol/kg. By mass: $2.69\times 58.44=157$ g of salt per kilogram of water, a solution of about 14% by mass. That is considerable, and it explains the amount of salt visible on roads in winter.

$\Delta T_{b}=i\,K_{b}\,m=2\times 0.512\times 2.00=2.05$ °C, so boiling occurs at $102.05$ °C. The effect is much smaller than the freezing-point effect, since $K_{b}$ is almost four times smaller than $K_{f}$. Salting pasta water therefore does NOT make it boil noticeably hotter: at a spoonful for two litres, the rise is a few hundredths of a degree.

Because salt reaches its physical limit: the water-$NaCl$ solution has a eutectic point at about $-21$ °C, a temperature below which no proportion of salt prevents freezing. Well before this threshold, effectiveness deteriorates and the quantities needed become unreasonable, as the calculation in c) shows. Cities then switch to calcium chloride, which releases three particles instead of two and whose dissolution is, moreover, exothermic, or to abrasives such as sand and gravel, which lower nothing but restore traction.

$\pi=iCRT$ with $i=1$ for glucose, a nonelectrolyte, and $T=310$ K: $\pi=1\times 0.100\times 8.314\times 310=258$ kPa, or about 2.5 atmospheres. A considerable pressure for such a dilute solution.

$NaCl$ releases two ions, so $i=2$: $\pi=2\times 0.154\times 8.314\times 310=794$ kPa. The value is very close to the 770 kPa of blood: this is precisely why physiological saline is prepared at this concentration, called ISOTONIC. It appears expressed as a mass percent in the familiar form “0.9% NaCl”.

With pure water, the external osmotic pressure would be zero while the inside of the red blood cells stays at 770 kPa: water would flow massively into the cells, which would swell and then BURST, a phenomenon called hemolysis. With a solution that is too concentrated, hypertonic, the movement would reverse: water would leave the cells, which would shrivel, what is called crenation. In both cases the cells are destroyed, which makes the choice of concentration literally vital.

From $\pi=CRT$ with $C=\dfrac{n}{V}=\dfrac{m}{MV}$, we get $M=\dfrac{mRT}{\pi V}=\dfrac{1.00\times 8.314\times 298}{2.50\times 0.100}=9.91\times 10^{3}$ g/mol, or about 9900 g/mol. The factor $i$ equals 1: a globular protein does not dissociate.

The molality of the solution is about $\dfrac{1.00/9910}{0.100}=1.01\times 10^{-3}$ mol/kg, so $\Delta T_{f}=1.86\times 1.01\times 10^{-3}=1.9\times 10^{-3}$ °C. Two thousandths of a degree: far below the precision of an ordinary laboratory thermometer, so the measurement is impossible in practice. The osmotic pressure, by contrast, reaches 2.50 kPa, or 25 cm of water column, perfectly readable on a simple graduated tube. CONCLUSION: for macromolecules, we measure the osmotic pressure; freezing-point depression can only be used for small molecules, whose molalities are a thousand times larger.

$n_{glucose}=\dfrac{45.0}{180.2}=0.2497$ mol and $n_{water}=\dfrac{250}{18.02}=13.87$ mol, for a total of 14.12 mol. The mole fraction of water is $x_{water}=\dfrac{13.87}{14.12}=0.9823$. The vapour pressure of the solution is therefore $P=0.9823\times 3.17=3.114$ kPa, and the lowering $\Delta P=3.17-3.114=0.056$ kPa. It is a small but perfectly measurable effect, and above all it depends only on the NUMBER of dissolved particles, never on their nature: that is what defines a colligative property.

Since $x_{solvent}+x_{solute}=1$, we write $\Delta P=P^{\circ}-P=P^{\circ}-x_{solvent}P^{\circ}=P^{\circ}\left(1-x_{solvent}\right)=x_{solute}P^{\circ}$. Numerically, $x_{solute}=\dfrac{0.2497}{14.12}=0.01768$ and $\Delta P=0.01768\times 3.17=0.0560$ kPa: we do recover the value of a). This form is more convenient because it gives the small quantity sought directly, whereas the first makes it appear as the difference of two nearby numbers, 3.17 and 3.114. If $x_{water}$ is rounded to 0.98 in the first method, the lowering comes out as 0.063 kPa, a 12% error; the second method does not suffer from this problem. It is a general rule of calculation: when you are looking for a small difference, try to express it directly.

Here both components are volatile, so each contributes to the pressure according to Raoult's law. With $x_{benzene}=x_{toluene}=0.500$: $P_{benzene}=0.500\times 12.7=6.35$ kPa and $P_{toluene}=0.500\times 3.79=1.895$ kPa. The total pressure is $P_{tot}=6.35+1.895=8.245$ kPa, or about 8.25 kPa. The composition of the vapour is obtained from Dalton's law, since the mole fractions in the gas phase are the ratios of the partial pressures to the total pressure: $y_{benzene}=\dfrac{6.35}{8.245}=0.770$ and $y_{toluene}=\dfrac{1.895}{8.245}=0.230$.

The liquid is 50% benzene, the vapour 77%: the vapour is clearly enriched in benzene, that is, in the more volatile component. This is logical, since the vapour pressure of pure benzene is 3.35 times that of toluene: at equal mole fraction in the liquid, it sends 3.35 times more molecules into the gas phase. If this vapour is condensed, we get a liquid that is 77% benzene; vaporizing it again, the new vapour will be richer still, and so on. A fractional distillation column is nothing other than the automatic repetition of this cycle over a large number of plates, which makes it possible to separate two liquids that simple boiling does not separate. Finally, the boiling-point elevation follows directly from a): since a nonvolatile solute lowers the vapour pressure, more heating is needed to bring this pressure back up to atmospheric pressure, and so to make the liquid boil. The freezing-point depression is explained in the same way on the solid side. The three colligative properties studied in this set are therefore just three consequences of a single fact: the solvent is diluted.

The sample contains $10.0\times 0.950=9.5$ g of benzoic acid and 0.5 g of impurities. At 100 °C, 100 mL dissolves 6.80 g, so we need $V=\dfrac{9.5}{6.80}\times 100=140$ mL of boiling water.

At 20 °C, these 140 mL can only hold $\dfrac{0.29}{100}\times 140=0.41$ g of benzoic acid. The rest crystallizes: $9.5-0.41=9.09$ g. The yield is $\dfrac{9.09}{9.5}\times 100=95.7\%$. The 0.41 g lost cannot be recovered this way: it is the unavoidable price of the method, since a saturated solution has to remain.

With 200 mL, the cold solution holds $\dfrac{0.29}{100}\times 200=0.58$ g, and only $9.5-0.58=8.92$ g crystallizes, a yield of 93.9%. The loss, 1.8 points, looks modest here, but it is directly proportional to the volume of water used: with a litre, we would lose 2.9 g, nearly a third of the product. The rule is therefore absolute in practice: use the MINIMUM volume of boiling solvent, added in small portions until dissolution is only just complete. Any excess solvent is paid for in lost product.

The impurities stay in solution because they are present in very small amounts, 0.5 g in 140 mL here: even when cold, they do not reach their solubility limit, so nothing forces them to crystallize. The main product, for its part, far exceeds its own and has no choice but to precipitate. The separation therefore does not rest on a difference in nature between product and impurities, but on their difference in amount, which explains why a recrystallization is very effective on an already fairly pure product and almost useless on a 50-50 mixture. A good recrystallization solvent must meet two conditions: dissolve the product very well when hot and very poorly when cold, the solubility gap being what creates the yield, and leave the impurities either very soluble when cold, so that they stay in the mother liquor, or completely insoluble when hot, so that they are removed by hot filtration. Finally, abrupt cooling makes a very large number of small crystals grow at once: they trap solvent and impurities in their defects and on their large surface. Slow cooling gives a few nuclei time to grow into large, well-ordered crystals, which naturally reject whatever does not fit in their lattice. We therefore recover a little less material, but much purer: in a purification, purity decides, not mass.

From $\Delta T_{f}=i\,K_{f}\,m$ we get $i=\dfrac{\Delta T_{f}}{K_{f}\,m}=\dfrac{0.348}{1.86\times 0.100}=1.87$. The theoretical value is 2: dissociation therefore appears to be incomplete by about 6%.

$i=\dfrac{0.225}{1.86\times 0.100}=1.21$, against a theoretical value of 2 as well. The gap here is much larger, nearly 40%.

Because of ION PAIRING. In solution, a cation and an anion can stay associated for the duration of a few collisions, forming a pair that behaves as ONE particle in the colligative sense. The strength of this association follows Coulomb's law, and so the product of the charges. For $NaCl$, the charges are $+1$ and $-1$, product 1. For $MgSO_{4}$, they are $+2$ and $-2$, product 4: the attraction is four times stronger, and the pairs far more numerous. The number of truly independent particles therefore drops further.

$i$ INCREASES toward its theoretical value on dilution. Indeed, when diluting, the ions are on average farther apart, the electrostatic attraction energy between two given ions decreases, and the pairs come apart. At infinite dilution, $i$ tends exactly to the theoretical value. This is why the colligative laws are stated as LIMITING laws, valid for dilute solutions, and why tables often give $i$ as a function of concentration.

The value $i=1.02$, barely above 1, reveals that acetic acid is a WEAK acid: it dissociates only very slightly, by a few percent, and therefore remains almost entirely in the molecular form $CH_{3}COOH$. Each dissolved molecule then counts as only one particle, hence $i\approx 1$. The colligative measurement thus provides an independent method for telling a strong electrolyte from a weak one, and even for calculating a degree of dissociation: here $\alpha\approx i-1=0.02$, or 2%, which agrees with the order of magnitude obtained from $K_{a}$ in the chapter on acids and bases.

Glycol is a nonelectrolyte, so $i=1$. From $\Delta T_{f}=K_{f}\,m$ we get $m=\dfrac{20.0}{1.86}=10.75$ mol/kg. For 5.00 kg of water: $n=10.75\times 5.00=53.8$ mol, a mass of $53.8\times 62.07=3.34\times 10^{3}$ g, about 3.34 kg.

The volume is $\dfrac{3337}{1.11}=3.01\times 10^{3}$ mL, or 3.01 L. The final mixture therefore contains 5.00 L of water and 3.01 L of glycol, which represents about 38% of the total volume. This matches the proportions of commercial antifreezes, generally sold ready to use at between 40 and 50% glycol.

$\Delta T_{b}=K_{b}\,m=0.512\times 10.75=5.51$ °C, so boiling occurs at about $105.5$ °C instead of 100 °C, at atmospheric pressure. This is a direct advantage: an engine runs at high temperature, and a coolant that boiled would form vapour bubbles, which insulate thermally and cause local overheating. Glycol therefore protects at BOTH ends of the scale, against freezing in winter and against boiling in summer. The circuit being under pressure as well, the real margin is greater still.

$\pi=CRT=1.10\times 8.314\times 298=2.73\times 10^{3}$ kPa, or about 27 atmospheres. To reverse the natural flow of osmosis and push pure water from the salty side to the desalinated side, the plant must apply a pressure HIGHER than this value, in practice 50 to 80 bar to obtain a usable flow. This is what makes desalination energy-intensive: we pay a pressure, and so a pumping work, for every litre produced.

For three reasons, two of them unrelated to colligative properties. First and above all, salt is extremely CORROSIVE to the metals of the circuit, aluminum cylinder head and copper radiator, which glycol is not. Next, the solubility of $NaCl$ levels off around 6 mol/kg, which, even with $i=2$, limits the depression to about 21 °C, corresponding to the eutectic point already met in exercise 5: it is impossible to go lower, whereas glycol, miscible in all proportions, allows it. Finally, glycol lubricates the water pump, which a brine would not. The chemistry of the chapter therefore gives half the answer; corrosion gives the other half.