1.3 Colligative Properties and Abnormal Molar Masses
Four properties of a dilute solution — the relative lowering of vapour pressure, the elevation of boiling point, the depression of freezing point and the osmotic pressure — depend only on how many solute particles are present. They measure molar masses, and the van 't Hoff factor $i$ corrects them for solutes that split up or pair up.
In Section 1.2 you saw that a non-volatile solute lowers the vapour pressure of a solvent by an amount that depends only on its mole fraction, $\Delta p_{1}=x_{2}\,p_{1}^{\circ}$. Glucose, urea and sucrose all have the same effect, molecule for molecule. That one fact has surprising consequences. Because the vapour pressure is lower, the solution must be heated more before it boils and cooled more before it freezes. And if the solution is separated from pure solvent by a suitable membrane, solvent flows into it, with enough push to raise a column of liquid metres high. These effects keep a car engine from freezing in a Ladakh winter, let a doctor choose a safe drip for a patient, and supply drinking water to coastal cities from the sea.
Properties that depend on the number of solute particles, and not on their identity, are called colligative (from the Latin colligare, to bind together). Count the particles and you can predict the property; measure the property and you can count the particles. Knowing the mass dissolved, you then get the molar mass. When the answer comes out "wrong", for example half the expected value for common salt, the solute is breaking up into ions or joining into pairs. The van 't Hoff factor puts that into numbers.
Colligative Properties and the Relative Lowering of Vapour Pressure
A colligative property of a dilute solution of a non-volatile solute depends only on the number of solute particles in a given amount of solvent, and not on the nature of those particles. There are four:
- the relative lowering of vapour pressure of the solvent, $\Delta p_{1}/p_{1}^{\circ}$;
- the elevation of the boiling point, $\Delta T_{b}$;
- the depression of the freezing point, $\Delta T_{f}$;
- the osmotic pressure, $\pi$.
The equations of this section hold for dilute ideal solutions of solutes that neither dissociate nor associate. The last part of the section shows how to correct them when the solute does either.
For a dilute solution of a non-volatile, non-electrolyte solute, the relative lowering of the vapour pressure of the solvent equals the mole fraction of the solute:
This is Result 1.5 of Section 1.2 divided by $p_{1}^{\circ}$. It does not depend on the solvent's own vapour pressure. For a very dilute solution $n_{2}\ll n_{1}$, so $x_{2}\approx n_{2}/n_{1}$, and with $n=w/M$ (mass $w$ in grams, molar mass $M$ in $\mathrm{g\,mol^{-1}}$):
Solving for $M_{2}$ gives the molar mass of the solute. The exact form, with $n_{1}+n_{2}$ in the denominator, should be used whenever $x_{2}$ is more than about $0.02$, or whenever the question gives enough data to do so.
Elevation of Boiling Point and Depression of Freezing Point
A liquid boils when its vapour pressure reaches the external pressure. Pure water reaches $1.013\ \mathrm{bar}$ at $373.15\ \mathrm{K}$. A solution has a lower vapour pressure at every temperature, so at $373.15\ \mathrm{K}$ it has not yet reached $1.013\ \mathrm{bar}$, and it must be heated a little more. Its boiling point is raised.
A liquid freezes when its vapour pressure equals that of the solid. Only pure solvent freezes out of a dilute solution (the solid is pure ice), so the solid's vapour-pressure curve is unchanged. The solution's curve is lower, and it meets the solid's curve at a lower temperature. Its freezing point is lowered. Interactive 1 draws both curves.
For a dilute solution of a non-volatile solute, the elevation of the boiling point $\Delta T_{b}=T_{b}-T_{b}^{\circ}$ is proportional to the molality $m$ of the solute:
($w_{1}$ and $w_{2}$ in grams). $K_{b}$, the ebullioscopic constant or molal elevation constant, is the elevation produced by a $1\ \mathrm{mol\,kg^{-1}}$ solution (extrapolated from dilute solutions). Its unit is $\mathrm{K\,kg\,mol^{-1}}$. It depends only on the solvent:
with $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $M_{1}$ in $\mathrm{g\,mol^{-1}}$ and $\Delta_{vap}H$ in $\mathrm{J\,mol^{-1}}$. The molality is used, not the molarity, because the measurement involves a change of temperature, and molality does not change with temperature (Section 1.1).
For a dilute solution from which pure solvent freezes out, the depression of the freezing point $\Delta T_{f}=T_{f}^{\circ}-T_{f}$ is proportional to the molality of the solute:
$K_{f}$, the cryoscopic constant or molal depression constant (unit $\mathrm{K\,kg\,mol^{-1}}$), depends only on the solvent:
Both $\Delta T_{b}$ and $\Delta T_{f}$ are written as positive numbers. A kelvin and a degree Celsius are the same size, so $\Delta T=0.52\ \mathrm{K}$ is also $0.52\ ^{\circ}\mathrm{C}$.
| Solvent | $T_{b}^{\circ}$ / K | $K_{b}$ / $\mathrm{K\,kg\,mol^{-1}}$ | $T_{f}^{\circ}$ / K | $K_{f}$ / $\mathrm{K\,kg\,mol^{-1}}$ |
|---|---|---|---|---|
| water | 373.15 | 0.52 | 273.15 | 1.86 |
| ethanol | 351.5 | 1.20 | 155.7 | 1.99 |
| benzene | 353.3 | 2.53 | 278.6 | 5.12 |
| carbon disulphide | 319.4 | 2.34 | 161.6 | 3.83 |
| cyclohexane | 353.9 | 2.79 | 279.6 | 20.0 |
| ethanoic acid | 391.1 | 3.07 | 289.8 | 3.90 |
Two patterns in the table are worth noticing. $K_{f}$ is larger than $K_{b}$ for most solvents, so freezing-point measurements give bigger, more accurate readings. And $K_{f}$ for cyclohexane is about four times that of benzene, which makes it a good solvent for finding molar masses. Check the formula for water: with $M_{1}=18.02$, $T_{b}^{\circ}=373.15\ \mathrm{K}$ and $\Delta_{vap}H=40.66\ \mathrm{kJ\,mol^{-1}}$,
$=0.513\ \mathrm{K\,kg\,mol^{-1}}$, close to the measured $0.52$. With $T_{f}^{\circ}=273.15\ \mathrm{K}$ and $\Delta_{fus}H=6.01\ \mathrm{kJ\,mol^{-1}}$ the same calculation gives $K_{f}=1.86\ \mathrm{K\,kg\,mol^{-1}}$.
Osmosis and Osmotic Pressure
Put a raisin in water overnight and it swells; sprinkle salt on a slice of cucumber and it goes limp as water seeps out. In both cases water moves through a cell wall from where it is more plentiful (fewer dissolved particles) to where it is less plentiful (more dissolved particles). The cell wall lets water through but holds back sugar or salt.
- A semipermeable membrane (SPM) allows solvent molecules to pass but not solute particles. Natural examples are animal bladders, parchment and the membranes of living cells; synthetic ones include cellophane and films of polymers or of copper(II) hexacyanidoferrate(II).
- Osmosis is the net flow of solvent through a semipermeable membrane from the pure solvent into a solution, or from a more dilute solution into a more concentrated one.
- The osmotic pressure $\pi$ of a solution is the extra pressure that must be applied to the solution to stop osmosis when it is separated from pure solvent by a semipermeable membrane.
Why does solvent flow at all? The solvent in the solution has a lower escaping tendency, which is the same fact that lowers its vapour pressure. Solvent molecules cross the membrane in both directions, but more cross from the pure side, where every molecule hitting the membrane is a solvent molecule. Pressing on the solution side raises the escaping tendency of its solvent until the two flows balance; the pressure needed is $\pi$. In an open U-tube, the liquid rises on the solution side until the extra hydrostatic pressure $h\rho g$ of the raised column equals $\pi$.
For a dilute solution of a non-electrolyte, the osmotic pressure is proportional to the molar concentration $C$ of the solute and to the absolute temperature:
where $V$ is the volume of the solution in litres. With $R=0.083\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$ and $C$ in $\mathrm{mol\,L^{-1}}$, $\pi$ comes out in bar. (With $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, use $C$ in $\mathrm{mol\,m^{-3}}$ to get pascals.) Because $n_{2}=w_{2}/M_{2}$,
The equation looks like the ideal gas law $pV=nRT$. Here the solute particles, spread through the solvent, behave like gas molecules in an otherwise empty container. Molarity can be used here because the measurement is made at a single, fixed temperature.
Why osmotic pressure is the method of choice for large molecules. The measurement is done at room temperature, so heat-sensitive proteins and polymers are not damaged. It uses molarity rather than molality, which is easy to prepare. Above all, the effect is large: a solution far too dilute to change the freezing point measurably still gives an osmotic pressure of a few centimetres of water, as Example 4 shows.
Two solutions with the same osmotic pressure at a given temperature are isotonic: when they are separated by a semipermeable membrane, there is no net osmosis. Relative to another solution, a solution of higher osmotic pressure is hypertonic and one of lower osmotic pressure is hypotonic.
The fluid inside red blood cells is isotonic with $0.9\%$ (w/v) sodium chloride ("normal saline"), so cells keep their shape in it. In a hypertonic solution (more than $0.9\%$) water leaves the cells and they shrink; in a hypotonic solution (less than $0.9\%$, or pure water) water enters, and the cells swell and may burst. This is why intravenous fluids must be isotonic with blood.
If a pressure larger than the osmotic pressure is applied to the solution side, the flow reverses: pure solvent is squeezed out of the solution through the membrane. This is reverse osmosis (RO). Seawater behaves roughly like $0.55\ \mathrm{mol\,L^{-1}}$ $\mathrm{NaCl}$, so its osmotic pressure is about $2\times0.55\times0.083\times298\approx27\ \mathrm{bar}$. Desalination plants, such as those that supply Chennai, push seawater against membranes of cellulose acetate or polyamide at about $55$–$70\ \mathrm{bar}$. The extra pressure keeps water flowing even as the brine left behind becomes saltier. Household RO purifiers work the same way on tap water, which has a far lower osmotic pressure.
Slips to avoid. (i) $\pi=CRT$ uses molarity and $T$ in kelvin. (ii) Match $R$ to the units: $0.083\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$ gives bar; $0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$ gives atm. (iii) Solvent flows towards the solution with more particles, never the other way, unless an external pressure greater than $\pi$ is applied. (iv) Count particles, not formula units, for electrolytes (next part).
Abnormal Molar Masses and the van 't Hoff Factor
Dissolve $0.100\ \mathrm{mol}$ of $\mathrm{NaCl}$ in $1\ \mathrm{kg}$ of water and the freezing point falls by about $0.35\ \mathrm{K}$, almost twice the $0.186\ \mathrm{K}$ expected. Work backwards with Law 1.8 and the "molar mass" of $\mathrm{NaCl}$ comes out near $31\ \mathrm{g\,mol^{-1}}$ instead of $58.5\ \mathrm{g\,mol^{-1}}$. The explanation is simple: $\mathrm{NaCl}$ gives $\mathrm{Na^{+}(aq)}$ and $\mathrm{Cl^{-}(aq)}$, so there are nearly twice as many particles as formula units. The opposite happens to ethanoic acid in benzene. Its molecules pair up through hydrogen bonds into dimers, $\mathrm{2CH_{3}COOH\rightleftharpoons(CH_{3}COOH)_{2}}$, so there are fewer particles than molecules, and the measured molar mass comes out near $120\ \mathrm{g\,mol^{-1}}$ instead of $60$.
A molar mass found from a colligative property that differs from the true molar mass, because the solute dissociates or associates, is an abnormal molar mass. The van 't Hoff factor $i$ measures the effect:
$i\gt1$ for dissociation, $i\lt1$ for association, and $i=1$ for a solute that does neither.
Dissociation. If one formula unit gives $\nu$ particles (e.g. $\nu=2$ for $\mathrm{KCl}$, $3$ for $\mathrm{K_{2}SO_{4}}$, $5$ for $\mathrm{Al_{2}(SO_{4})_{3}}$) and a fraction $\alpha$ of the solute dissociates, then 1 mol gives $(1-\alpha)+\nu\alpha$ mol of particles:
Association. If $n$ molecules join into one ($n=2$ for dimers) and a fraction $\beta$ of the solute associates, then 1 mol gives $(1-\beta)+\beta/n$ mol of particles:
For complete dissociation of a strong electrolyte the ideal value is $i=\nu$. Measured values are a little smaller, and approach $\nu$ only at high dilution, because oppositely charged ions attract one another and do not move fully independently.
(b) $2.00\ \mathrm{g}$ of ethanoic acid ($\mathrm{CH_{3}COOH}$, $60.0\ \mathrm{g\,mol^{-1}}$) in $100\ \mathrm{g}$ of benzene lowers the freezing point by $0.905\ \mathrm{K}$ ($K_{f}=5.12\ \mathrm{K\,kg\,mol^{-1}}$). Find the apparent molar mass, $i$, and the percentage of the acid that is dimerised.
1. Molality vs molarity. $\Delta T_{b}$ and $\Delta T_{f}$ use molality (per kg of solvent); $\pi$ uses molarity (per litre of solution). In very dilute aqueous solutions the two are nearly equal, which is why exam questions sometimes mix them.
2. Forgetting $i$. $0.1\ \mathrm{m}$ $\mathrm{CaCl_{2}}$ has up to $0.3\ \mathrm{mol\,kg^{-1}}$ of particles. When comparing boiling or freezing points, compare $i\times m$, not $m$.
3. Signs and the final answer. $\Delta T_{f}$ is a positive depression; the new freezing point is $T_{f}^{\circ}-\Delta T_{f}$. Questions often ask for the freezing point itself, not the depression.
4. Using the solution mass for $w_{1}$. In $m=\dfrac{1000\,w_{2}}{M_{2}\,w_{1}}$, $w_{1}$ is the mass of the solvent only (Section 1.1).
Exercises
- 1 What is a colligative property? Name the four colligative properties and explain why they are "colligative".
- 2 $6.00\ \mathrm{g}$ of urea is dissolved in $90.0\ \mathrm{g}$ of water at $298\ \mathrm{K}$. Find the relative lowering of vapour pressure and the vapour pressure of the solution.
- 3 The relative lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte is $0.0125$. Find the molality of the solution.
- 4 Find the boiling point of a $0.250\ \mathrm{mol\,kg^{-1}}$ aqueous sucrose solution at $1\ \mathrm{atm}$.
- 5 $2.50\ \mathrm{g}$ of a non-volatile non-electrolyte in $75.0\ \mathrm{g}$ of benzene raises the boiling point by $0.330\ \mathrm{K}$. Find its molar mass.
- 6 Calculate $K_{b}$ for ethanol from $\Delta_{vap}H=38.6\ \mathrm{kJ\,mol^{-1}}$, $T_{b}^{\circ}=351.5\ \mathrm{K}$ and $M_{1}=46.07\ \mathrm{g\,mol^{-1}}$, and compare it with the table.
- 7 Which boils at the higher temperature: $0.10\ \mathrm{mol\,kg^{-1}}$ aqueous glucose or $0.10\ \mathrm{mol\,kg^{-1}}$ aqueous urea? Explain.
- 8 Kitchen chemistry. A cook adds $10.0\ \mathrm{g}$ of $\mathrm{NaCl}$ to $1.00\ \mathrm{kg}$ of water for boiling pasta. Assuming $i=2$, by how much does the boiling point rise? Does salt make pasta cook noticeably faster?
- 1 Find the freezing point of a $0.400\ \mathrm{mol\,kg^{-1}}$ aqueous urea solution.
- 2 $1.00\ \mathrm{g}$ of a non-electrolyte in $50.0\ \mathrm{g}$ of benzene lowers the freezing point by $0.640\ \mathrm{K}$. Find its molar mass.
- 3 A $4.00\%$ (by mass) aqueous sucrose solution freezes $0.227\ \mathrm{K}$ below pure water. Without using $K_{f}$, predict the freezing-point depression of a $4.00\%$ glucose solution. Then check your answer with $K_{f}$.
- 4 What mass of methanol must be added to $1.00\ \mathrm{kg}$ of water to keep it from freezing down to $-15.0\ ^{\circ}\mathrm{C}$? Why is methanol a poor choice for a car radiator?
- 5 Road salt. Estimate the freezing point of a $10.0\%$ (by mass) $\mathrm{NaCl}$ solution, assuming complete dissociation into two ions.
- 6 $0.500\ \mathrm{g}$ of naphthalene in $25.0\ \mathrm{g}$ of benzene lowers the freezing point by $0.800\ \mathrm{K}$. Find $K_{f}$ for benzene.
- 7 An aqueous solution of a non-electrolyte boils at $100.26\ ^{\circ}\mathrm{C}$ at $1\ \mathrm{atm}$. At what temperature does it freeze?
- 1 Find the osmotic pressure of a $0.0500\ \mathrm{mol\,L^{-1}}$ glucose solution at $310\ \mathrm{K}$.
- 2 $1.00\ \mathrm{g}$ of urea is dissolved to make $200\ \mathrm{mL}$ of solution at $298\ \mathrm{K}$. Find its osmotic pressure.
- 3 $0.800\ \mathrm{g}$ of a synthetic polymer in $100\ \mathrm{mL}$ of solution at $300\ \mathrm{K}$ has an osmotic pressure of $1.66\times10^{-3}\ \mathrm{bar}$. Find its molar mass.
- 4 Which of these solutions are isotonic with one another at the same temperature: $0.10\ \mathrm{M}$ glucose, $0.050\ \mathrm{M}$ $\mathrm{NaCl}$, $0.10\ \mathrm{M}$ urea, $0.050\ \mathrm{M}$ glucose? (Take $i=2$ for $\mathrm{NaCl}$.)
- 5 Red blood cells are placed in (a) $0.50\%$ and (b) $2.0\%$ (w/v) $\mathrm{NaCl}$. Describe and explain what happens in each case.
- 6 Desalination. Treat seawater as $0.55\ \mathrm{mol\,L^{-1}}$ $\mathrm{NaCl}$ with $i=2$. Find its osmotic pressure at $298\ \mathrm{K}$. What is the least pressure an RO plant must apply, and why do real plants use much more?
- 7 Find the osmotic pressure of $0.010\ \mathrm{mol\,L^{-1}}$ sucrose at $300\ \mathrm{K}$, and the height of a water column ($\rho=1000\ \mathrm{kg\,m^{-3}}$, $g=9.81\ \mathrm{m\,s^{-2}}$) that it can support.
- 8 Give two reasons why osmotic pressure, rather than freezing-point depression, is used to find the molar masses of proteins.
- 1 Give the ideal van 't Hoff factor for complete dissociation or association of: (a) $\mathrm{KCl}$; (b) $\mathrm{K_{2}SO_{4}}$; (c) $\mathrm{Al_{2}(SO_{4})_{3}}$; (d) glucose; (e) ethanoic acid fully dimerised in benzene.
- 2 Find the freezing-point depression of $0.0100\ \mathrm{mol\,kg^{-1}}$ aqueous $\mathrm{K_{2}SO_{4}}$, assuming complete dissociation.
- 3 A $0.050\ \mathrm{mol\,kg^{-1}}$ aqueous $\mathrm{MgCl_{2}}$ solution freezes at $-0.250\ ^{\circ}\mathrm{C}$. Find $i$ and the apparent degree of dissociation.
- 4 A $0.0100\ \mathrm{mol\,L^{-1}}$ solution of a salt $\mathrm{MX}$ has an osmotic pressure of $0.465\ \mathrm{bar}$ at $300\ \mathrm{K}$. Find $i$ and $\alpha$.
- 5 $2.44\ \mathrm{g}$ of benzoic acid in $100\ \mathrm{g}$ of benzene lowers the freezing point by $0.533\ \mathrm{K}$. Find the apparent molar mass and the percentage of the acid present as dimers.
- 6 Arrange $0.1\ \mathrm{mol\,kg^{-1}}$ aqueous solutions of glucose, $\mathrm{NaCl}$, $\mathrm{CaCl_{2}}$ and $\mathrm{Al_{2}(SO_{4})_{3}}$ in order of increasing freezing point (assume complete dissociation).
- 7 $0.585\ \mathrm{g}$ of $\mathrm{NaCl}$ in $100\ \mathrm{g}$ of water lowers the freezing point by $0.344\ \mathrm{K}$. Find the apparent molar mass of $\mathrm{NaCl}$, $i$ and $\alpha$.
E1. Board · 3 marks (a) Define osmotic pressure. (b) Why is osmotic pressure preferred over other colligative properties for finding the molar masses of macromolecules? (c) Calculate the osmotic pressure at $300\ \mathrm{K}$ of a solution made by dissolving $1.80\ \mathrm{g}$ of glucose ($180\ \mathrm{g\,mol^{-1}}$) in water to give $500\ \mathrm{mL}$ of solution ($R=0.083\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$).
E2. JEE Main Numerical value. A weak monobasic acid is $10\%$ ionised in a $0.20\ \mathrm{mol\,kg^{-1}}$ aqueous solution. Taking $K_{f}=1.86\ \mathrm{K\,kg\,mol^{-1}}$, the depression of freezing point is $x\times10^{-2}\ \mathrm{K}$. The value of $x$, to the nearest integer, is ______.
E3. NEET Which $0.1\ \mathrm{mol\,kg^{-1}}$ aqueous solution has the highest boiling point? (A) glucose (B) $\mathrm{NaCl}$ (C) $\mathrm{BaCl_{2}}$ (D) urea
E4. JEE Advanced $1.20\ \mathrm{g}$ of ethanoic acid ($60.0\ \mathrm{g\,mol^{-1}}$) is dissolved in $50.0\ \mathrm{g}$ of benzene ($K_{f}=5.12\ \mathrm{K\,kg\,mol^{-1}}$, $T_{f}^{\circ}=278.6\ \mathrm{K}$), and the solution freezes at $277.37\ \mathrm{K}$. The acid partly forms dimers. (a) Find the van 't Hoff factor and the percentage of acid molecules that are dimerised (nearest integer). (b) If the same mass of acid were dissolved in $50.0\ \mathrm{g}$ of water, where it is only slightly ionised, would the freezing-point depression be larger or smaller than "normal"? Explain.
E5. NEET Board Assertion–reason. Assertion (A): The depression of freezing point of $0.01\ \mathrm{mol\,kg^{-1}}$ aqueous $\mathrm{NaCl}$ is nearly twice that of $0.01\ \mathrm{mol\,kg^{-1}}$ aqueous glucose. Reason (R): $\mathrm{NaCl}$ dissociates almost completely into $\mathrm{Na^{+}}$ and $\mathrm{Cl^{-}}$ ions in dilute aqueous solution. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
E6. Board · Case study A coastal town plans a reverse-osmosis plant. Its seawater behaves like $0.60\ \mathrm{mol\,L^{-1}}$ $\mathrm{NaCl}$ with an effective van 't Hoff factor of $1.9$ at $298\ \mathrm{K}$. (a) Explain what reverse osmosis is and why the pressure applied must exceed the osmotic pressure of the seawater. (2 marks) (b) Calculate that osmotic pressure ($R=0.083\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$). (1 mark) (c) Name one material used for RO membranes. (1 mark)
Research question. Oral rehydration solution (ORS) has saved millions of children with diarrhoea. Why did the World Health Organization lower its total particle concentration, and how does that choice compare with the osmotic pressure of blood?
Why it matters. In diarrhoea the body loses water and salts fast. A drink with the right balance of glucose and salts lets the gut absorb water again, by osmosis and by glucose-assisted uptake of sodium. If the drink is too concentrated (hypertonic), it can draw water into the gut and make things worse. India distributes ORS packets through every primary health centre.
What to do.
- The reduced-osmolarity ORS recipe per litre is: $\mathrm{NaCl}$ $2.6\ \mathrm{g}$, anhydrous glucose $13.5\ \mathrm{g}$, $\mathrm{KCl}$ $1.5\ \mathrm{g}$, trisodium citrate dihydrate ($\mathrm{Na_{3}C_{6}H_{5}O_{7}\cdot2H_{2}O}$, $294\ \mathrm{g\,mol^{-1}}$) $2.9\ \mathrm{g}$. Assuming complete dissociation of the salts, find the total concentration of particles in $\mathrm{mmol\,L^{-1}}$ ($\mathrm{NaCl}$ $58.5$, glucose $180$, $\mathrm{KCl}$ $74.6\ \mathrm{g\,mol^{-1}}$).
- Compute its osmotic pressure at body temperature ($310\ \mathrm{K}$) with $\pi=CRT$.
- Blood plasma has about $290\ \mathrm{mmol}$ of dissolved particles per kg of water. Compute its osmotic pressure at $310\ \mathrm{K}$ (take molality $\approx$ molarity). Is ORS isotonic, hypotonic or hypertonic to plasma?
- The older WHO formula had about $311\ \mathrm{mmol\,L^{-1}}$ of particles. Find out from WHO or UNICEF sources why the change was made, and what clinical trials showed.
Tools: Calculator or spreadsheet; Interactive 2 (set the molarity to your total particle concentration and choose glucose, $i=1$).
Hand in: A half-page note with your table of particles, the two osmotic pressures, and a two-sentence explanation of why a slightly hypotonic drink helps the gut absorb water.
Going further. Home-made ORS is often described as "six level teaspoons of sugar and half a teaspoon of salt in a litre of water". Estimate its particle concentration (a level teaspoon is about $4\ \mathrm{g}$ of sugar or $3\ \mathrm{g}$ of salt) and compare it with the WHO formula.
🧪 Self-check quiz
Q1. A colligative property of a dilute solution depends on
Q2. The freezing point of a $0.50\ \mathrm{mol\,kg^{-1}}$ aqueous urea solution ($K_{f}=1.86\ \mathrm{K\,kg\,mol^{-1}}$) is
Q3. Red blood cells are placed in $0.5\%$ (w/v) $\mathrm{NaCl}$ solution. They
Q4. The ideal van 't Hoff factor for $\mathrm{Al_{2}(SO_{4})_{3}}$ in very dilute aqueous solution is
Q5. Reverse osmosis takes place when