1.1 Types of Solutions, Concentration and Solubility
How much solute is in how much solution? Percentages, ppm, mole fraction, molarity and molality answer that question in different ways, and Henry's law $p=K_{H}\,x$ tells us how much gas a liquid can hold.
A hospital drip bag is labelled "sodium chloride 0.9% w/v". A car's coolant is "40% antifreeze by mass". A water report gives "fluoride 1.2 ppm", and a laboratory bottle says "$0.1\ \mathrm{M}$ $\mathrm{HCl}$". Each label describes the same idea — how much solute is mixed into how much solution — but in a different unit, chosen for a different job. A nurse cares about grams per volume, a physical chemist about moles per kilogram, an environmental scientist about very small mass ratios. You must be able to read every one of these labels and convert between them quickly and without error. Board papers, JEE and NEET all test exactly this.
The second half of the section asks a different question: how much of a substance can dissolve? For solids, temperature matters most. For gases, pressure matters most. A sealed bottle of soda water is fizzy because carbon dioxide was forced in at high pressure, and a diver who comes up too fast gets "the bends" because nitrogen comes out of the blood as bubbles. One simple law, Henry's law, explains both, and you will use it to calculate how much oxygen a lake can hold for its fish.
What Is a Solution?
When sugar dissolves in tea, the sugar molecules spread out evenly between the water molecules. You cannot see separate sugar particles, and every spoonful tastes equally sweet. That evenness is the key feature of a solution.
A solution is a homogeneous mixture of two or more substances. Its composition and properties are the same throughout, down to the molecular scale. A solution with two components is a binary solution.
The component present in the largest amount usually decides the physical state of the solution and is called the solvent. The other components are solutes. In this chapter we label the solvent $1$ and the solute $2$. So $n_{1}$, $x_{1}$ and $M_{1}$ refer to the solvent, and $n_{2}$, $x_{2}$ and $M_{2}$ refer to the solute.
Solute and solvent can each be a gas, a liquid or a solid. The solution takes the state of the solvent, so a binary solution can be of $3\times3=9$ types:
| Type of solution | Solute | Solvent | Everyday example |
|---|---|---|---|
| Gaseous | gas | gas | air: oxygen and argon in nitrogen |
| liquid | gas | chloroform vapour mixed with nitrogen | |
| solid | gas | camphor (or naphthalene) vapour in air | |
| Liquid | gas | liquid | carbon dioxide in soda water; oxygen in a lake |
| liquid | liquid | ethanol in water | |
| solid | liquid | glucose or common salt in water | |
| Solid | gas | solid | hydrogen dissolved in palladium |
| liquid | solid | amalgam of mercury with sodium | |
| solid | solid | brass (zinc in copper); copper in gold jewellery |
Most of Class 12 chemistry happens in liquid solutions, usually with water as the solvent (aqueous solutions). The rest of this chapter is about these.
Expressing Concentration
The concentration of a solution tells us how much solute is present in a given amount of solution or solvent. There are two families of units. The first uses only masses and volumes, which you can measure directly with a balance or a measuring cylinder. The second counts particles in moles, which is what matters in reactions and in the properties you will meet in Sections 1.2 and 1.3.
- Mass percentage (w/w): $\ \text{mass \% of a component}=\dfrac{\text{mass of the component}}{\text{total mass of the solution}}\times100$.
- Volume percentage (v/v), used for liquid–liquid solutions: $\ \dfrac{\text{volume of the component}}{\text{total volume of the solution}}\times100$.
- Mass by volume percentage (w/v), common in medicine and pharmacy: $\ \dfrac{\text{mass of solute in g}}{\text{volume of solution in mL}}\times100$, i.e. grams of solute per $100\ \mathrm{mL}$ of solution.
- Parts per million (ppm), for traces: $\ \dfrac{\text{number of parts of the component}}{\text{total number of parts of all components}}\times10^{6}$. The parts may be counted by mass, by volume or in moles, so always say which. For dilute aqueous solutions ($d\approx1.00\ \mathrm{g\,mL^{-1}}$), $1\ \text{ppm by mass}\approx1\ \mathrm{mg\,L^{-1}}$.
(b) A hand sanitiser is made by mixing $70.0\ \mathrm{mL}$ of ethanol with enough water to give exactly $100.0\ \mathrm{mL}$ of solution. What is its volume percentage of ethanol?
- Mole fraction of component $i$: $\ x_{i}=\dfrac{n_{i}}{\sum n}$. For a binary solution $x_{1}=\dfrac{n_{1}}{n_{1}+n_{2}}$, $x_{2}=\dfrac{n_{2}}{n_{1}+n_{2}}$, and always $x_{1}+x_{2}=1$. It has no unit.
- Molarity: $\ M=\dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$, unit $\mathrm{mol\,L^{-1}}$, written $\mathrm{M}$ ("molar").
- Molality: $\ m=\dfrac{\text{moles of solute}}{\text{mass of solvent in kilograms}}$, unit $\mathrm{mol\,kg^{-1}}$, written $\mathrm{m}$ ("molal").
Notice the two different denominators: molarity divides by the volume of the solution, while molality divides by the mass of the solvent only.
Temperature, Conversions and Choosing a Unit
When a solution warms up, it expands. The moles of solute and the masses of solute and solvent stay the same, but the volume grows. So any unit with a volume in it changes with temperature: molarity, w/v % and v/v % all decrease slightly as the solution is heated. Units built only from masses or moles do not change: mass %, ppm by mass, mole fraction and molality.
Let the solution have density $d$ (in $\mathrm{g\,mL^{-1}}$), and let the solute and solvent have molar masses $M_{2}$ and $M_{1}$ (in $\mathrm{g\,mol^{-1}}$). Let $w$ be the mass fraction of solute, so $w=\text{mass \%}/100$. Then
$$M=\frac{1000\,d\,w}{M_{2}},\qquad m=\frac{1000\,M}{1000\,d-M\,M_{2}},\qquad x_{2}=\frac{m\,M_{1}}{1000+m\,M_{1}}.$$Why. One litre of solution has mass $1000d$ grams. Of this, $wd\times1000$ grams is solute, which is $1000dw/M_{2}$ moles; that gives $M$. The solute in that litre weighs $M\,M_{2}$ grams, so the solvent weighs $1000d-M\,M_{2}$ grams; that gives $m$. Finally, $1000\ \mathrm{g}$ of solvent contains $m$ mol of solute and $1000/M_{1}$ mol of solvent, which gives $x_{2}$. Each formula holds for any binary solution, provided $d$ is measured at the temperature you are interested in.
- Choose a basis that matches the unit you are given: $100\ \mathrm{g}$ of solution for a mass %, $1\ \mathrm{L}$ of solution for a molarity, $1\ \mathrm{kg}$ of solvent for a molality, $1\ \mathrm{mol}$ in total for a mole fraction.
- Fill in a small table with the mass, moles and volume of the solute, the solvent and the solution for that basis. Use the molar mass to switch between grams and moles. Use the density to switch between the mass and the volume of the solution. Never use the density to find the volume of the solvent.
- Read off the unit you want straight from its definition.
- Check: $x_{1}+x_{2}=1$; for dilute aqueous solutions $M\approx m$; the volume of a solution is never simply the volume of the solvent.
1. Using the solvent volume for molarity. "$0.5\ \mathrm{mol}$ in $1\ \mathrm{L}$ of water" is not $0.5\ \mathrm{M}$. Molarity uses the final volume of the solution, which is why volumetric flasks are filled "up to the mark".
2. Using the solution mass for molality. The denominator of $m$ is the mass of the solvent only. In Example 4, dividing by $0.100\ \mathrm{kg}$ instead of $0.0630\ \mathrm{kg}$ gives $3.78$ instead of $5.99\ \mathrm{mol\,kg^{-1}}$.
3. Grams and kilograms, mL and L. $m$ needs kilograms and $M$ needs litres. Most wrong answers in exams are off by exactly $1000$.
4. Thinking $1\ \mathrm{M}=1\ \mathrm{m}$. They are nearly equal only for dilute aqueous solutions, where $d\approx1\ \mathrm{g\,mL^{-1}}$ and the solute mass is small.
Solubility of Solids and Gases; Henry's Law
Keep adding salt to water while stirring and eventually no more dissolves: crystals stay at the bottom however long you wait. This does not mean dissolving has stopped. At the molecular level, ions still leave the crystal surface and other ions still return to it, at equal rates. The solution is in dynamic equilibrium with the undissolved solid: $$\text{solute (undissolved)}\ \rightleftharpoons\ \text{solute (in solution)}.$$
Like dissolves like. A solute dissolves well when the attractions it forms with the solvent are similar in strength to those already present in the pure substances. Ionic and polar solutes ($\mathrm{NaCl}$, sugar) dissolve in polar solvents such as water. Non-polar solutes (naphthalene, anthracene) dissolve in non-polar solvents such as benzene.
Temperature and solids. In a nearly saturated solution, the dissolution equilibrium follows Le Chatelier's principle. If dissolving absorbs heat ($\Delta_{sol}H\gt0$, e.g. $\mathrm{KNO_{3}}$, $\mathrm{NH_{4}Cl}$), raising the temperature increases the solubility. If dissolving releases heat ($\Delta_{sol}H\lt0$, e.g. cerium(III) sulphate), raising the temperature decreases it. For $\mathrm{NaCl}$, $\Delta_{sol}H$ is small, so its solubility hardly changes with temperature (about $36\ \mathrm{g}$ per $100\ \mathrm{g}$ of water at room temperature).
Pressure and solids. Solids and liquids are almost incompressible, so pressure has no significant effect on the solubility of a solid in a liquid.
Gases. Gases are different. A dissolved gas is in equilibrium with the same gas above the liquid. Squeezing the gas into a smaller volume crowds more molecules against the surface. More of them enter the liquid each second until a new equilibrium is reached with more gas dissolved. The quantitative statement is Henry's law (William Henry, 1803).
At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. In terms of the mole fraction $x$ of the gas in the solution, $$p=K_{H}\,x,$$ where $p$ is the partial pressure of the gas in the vapour phase and $K_{H}$ is the Henry's law constant. Its value depends on the gas, the solvent and the temperature, and it has the units of pressure (often $\mathrm{kbar}$).
Conditions. The law works best at moderate pressures, for dilute solutions, and for gases that do not react with or ionise in the solvent. It fails for $\mathrm{NH_{3}}$ or $\mathrm{HCl}$ in water, and it is only approximate for $\mathrm{CO_{2}}$ in water, which partly forms $\mathrm{H_{2}CO_{3}}$. In a gas mixture, each gas obeys the law separately with its own partial pressure.
Reading $K_{H}$. For a fixed $p$, $x=p/K_{H}$. So the larger $K_{H}$, the lower the solubility. For most gases $K_{H}$ increases with temperature, so gases become less soluble as the liquid warms up.
| Gas (in water) | $T$ / K | $K_{H}$ / kbar | Gas (in water) | $T$ / K | $K_{H}$ / kbar |
|---|---|---|---|---|---|
| He | 293 | 144.97 | $\mathrm{O_{2}}$ | 293 | 34.86 |
| $\mathrm{H_{2}}$ | 293 | 69.16 | $\mathrm{O_{2}}$ | 303 | 46.82 |
| $\mathrm{N_{2}}$ | 293 | 76.48 | Ar | 298 | 40.3 |
| $\mathrm{N_{2}}$ | 303 | 88.84 | $\mathrm{CO_{2}}$ | 298 | 1.67 |
The table is used in the examples, exercises and Interactive 2. $\mathrm{CO_{2}}$ has by far the smallest $K_{H}$, so it is by far the most soluble gas listed. Its solubility is high partly because it reacts slightly with water.
Scuba diving and the bends. Under water, pressure increases by about $1\ \mathrm{bar}$ for every $10\ \mathrm{m}$ of depth. A diver breathing compressed air at $30\ \mathrm{m}$ (about $4\ \mathrm{bar}$) dissolves about four times as much nitrogen in the blood as at the surface. A fast ascent lowers the pressure suddenly. The excess nitrogen then comes out of solution as bubbles in the blood vessels and joints, a painful and dangerous condition called the bends. To lower this risk, tanks for deep dives use mixtures in which part of the nitrogen is replaced by helium, which is much less soluble ($K_{H}$ about twice that of $\mathrm{N_{2}}$). A slow, staged ascent is also essential.
High altitude. At high altitude the total pressure, and hence $p_{\mathrm{O_{2}}}$, is low. Less oxygen dissolves in the blood and tissues, and climbers feel weak and cannot think clearly, a condition called anoxia.
Slips to avoid. (i) Use the partial pressure of the gas, not the total pressure. (ii) Use the same pressure unit for $p$ and $K_{H}$ ($1\ \mathrm{kbar}=1000\ \mathrm{bar}$). (iii) A larger $K_{H}$ means a less soluble gas. (iv) $x$ is a mole fraction, not a molarity. Convert it with $n_{\text{gas}}\approx x\,n_{1}$, and remember that $1\ \mathrm{L}$ of water is about $55.6\ \mathrm{mol}$.
Exercises
- 1 Name the type of solution (state of solute, state of solvent) for: (a) brass; (b) camphor vapour in nitrogen; (c) soda water; (d) hydrogen in palladium.
- 2 $12.0\ \mathrm{g}$ of glucose is dissolved in $188\ \mathrm{g}$ of water. Find the mass percentage of glucose.
- 3 $25.0\ \mathrm{mL}$ of ethanol is diluted with water to $200\ \mathrm{mL}$ of solution. Find the volume percentage of ethanol.
- 4 $4.50\ \mathrm{g}$ of $\mathrm{NaOH}$ is dissolved and made up to $250\ \mathrm{mL}$ of solution. Find the mass by volume percentage.
- 5 A $1.5\ \mathrm{kg}$ water sample contains $0.45\ \mathrm{mg}$ of arsenic. Express this in ppm and in ppb (parts per billion) by mass.
- 6 What masses of $\mathrm{NaCl}$ and water are needed to make $500\ \mathrm{g}$ of a $0.90\%$ (by mass) saline solution?
- 7 Environment. A lake has $8.0\ \text{ppm}$ by mass of dissolved oxygen. What mass of $\mathrm{O_{2}}$ is dissolved in $1.0\ \mathrm{m^{3}}$ of its water (mass $1000\ \mathrm{kg}$)?
- 8 $300\ \mathrm{g}$ of a $25.0\%$ (by mass) sugar solution is mixed with $400\ \mathrm{g}$ of a $40.0\%$ solution. Find the mass percentage of sugar in the mixture.
- 1 Find the molarity of a solution containing $5.85\ \mathrm{g}$ of $\mathrm{NaCl}$ in $500\ \mathrm{mL}$ of solution.
- 2 What mass of urea is needed to make $250\ \mathrm{mL}$ of a $0.400\ \mathrm{M}$ solution?
- 3 Find the molality of $18.0\ \mathrm{g}$ of glucose dissolved in $500\ \mathrm{g}$ of water.
- 4 $46.0\ \mathrm{g}$ of ethanol is mixed with $36.0\ \mathrm{g}$ of water. Find the mole fraction of each.
- 5 What volume of $6.0\ \mathrm{M}$ $\mathrm{HCl}$ must be diluted to make $500\ \mathrm{mL}$ of $0.30\ \mathrm{M}$ $\mathrm{HCl}$?
- 6 $200\ \mathrm{mL}$ of $0.30\ \mathrm{M}$ $\mathrm{NaCl}$ is mixed with $300\ \mathrm{mL}$ of $0.20\ \mathrm{M}$ $\mathrm{NaCl}$. Find the molarity of the mixture (assume the volumes add).
- 7 Find the mole fraction of the solute in a $1.00\ \mathrm{mol\,kg^{-1}}$ aqueous solution.
- 8 The mole fraction of urea in an aqueous solution is $0.100$. Find its molality.
- 9 What is the molarity of pure water at a density of $1.00\ \mathrm{g\,mL^{-1}}$?
- 10 Concentrated hydrochloric acid is $36.5\%$ $\mathrm{HCl}$ by mass, with density $1.18\ \mathrm{g\,mL^{-1}}$. Find its molarity, molality and the mole fraction of $\mathrm{HCl}$.
- 11 Challenge. A $2.50\ \mathrm{mol\,kg^{-1}}$ aqueous glucose solution has density $1.15\ \mathrm{g\,mL^{-1}}$. Find its molarity.
- 1 The dissolution of $\mathrm{KNO_{3}}$ in water absorbs heat, while that of cerium(III) sulphate releases heat. Predict how each solubility changes on heating, and justify your answer.
- 2 Why does pressure have no noticeable effect on the solubility of a solid in a liquid, but a large effect on the solubility of a gas?
- 3 Find the mole fraction of $\mathrm{N_{2}}$ in water at $293\ \mathrm{K}$ in equilibrium with air in which $p_{\mathrm{N_{2}}}=0.78\ \mathrm{bar}$.
- 4 What partial pressure of $\mathrm{CO_{2}}$ at $298\ \mathrm{K}$ gives a mole fraction of $2.0\times10^{-3}$ of $\mathrm{CO_{2}}$ in water?
- 5 Arrange He, $\mathrm{N_{2}}$, $\mathrm{O_{2}}$ and Ar in order of increasing solubility in water at the same partial pressure, using the table values.
- 6 At a fixed temperature a gas has $x=4.0\times10^{-5}$ in water when its partial pressure is $2.0\ \mathrm{bar}$. Find $K_{H}$, and $x$ at $5.0\ \mathrm{bar}$.
- 7 How many millimoles, and how many milligrams, of $\mathrm{O_{2}}$ dissolve in $1.00\ \mathrm{L}$ of water at $293\ \mathrm{K}$ in contact with pure oxygen at $1.00\ \mathrm{bar}$?
- 8 Mountaineering. At about $5000\ \mathrm{m}$ altitude the air pressure is about $0.54\ \mathrm{bar}$, compared with $1.01\ \mathrm{bar}$ at sea level. Air has the same composition at both places. At the same temperature, what percentage of the sea-level amount of oxygen can dissolve in water there?
- 9 Using $K_{H}(\mathrm{O_{2}})$ at $293$ and $303\ \mathrm{K}$, by what percentage does the solubility of oxygen in water fall when the water warms from $20\ ^{\circ}\mathrm{C}$ to $30\ ^{\circ}\mathrm{C}$ (same $p_{\mathrm{O_{2}}}$)?
- 1 Explain why molality, not molarity, is used when a solution's properties are measured over a range of temperatures (as in Section 1.3).
- 2 A student claims that $1.00\ \mathrm{M}$ and $1.00\ \mathrm{mol\,kg^{-1}}$ aqueous $\mathrm{NaCl}$ are the same solution. Using $d=1.04\ \mathrm{g\,mL^{-1}}$ for the $1.00\ \mathrm{M}$ solution, find its molality and comment.
- 3 Starting from the definitions, show that for a binary solution with solvent molar mass $M_{1}$ (in $\mathrm{g\,mol^{-1}}$), $x_{2}=\dfrac{m\,M_{1}}{1000+m\,M_{1}}$, and that $x_{2}\approx\dfrac{m\,M_{1}}{1000}$ for dilute solutions.
- 4 Medicine. An intravenous bag holds $1.00\ \mathrm{L}$ of $5.0\%$ (w/v) glucose solution. Find the molarity of glucose and the mass of glucose a patient receives from three such bags.
- 5 Diving. At the same partial pressure and $293\ \mathrm{K}$, what is the ratio of the solubility (mole fraction) of helium to that of nitrogen in water? Explain why this makes helium useful in breathing gas for deep dives.
E1. Board (a) Define molality. (b) Why is molality preferred to molarity for expressing concentration in physical measurements? (c) Calculate the molality of a solution made by dissolving $3.00\ \mathrm{g}$ of urea ($M=60\ \mathrm{g\,mol^{-1}}$) in $150\ \mathrm{g}$ of water. (3 marks)
E2. JEE Main Numerical value. A $2.00\ \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ ($M=40\ \mathrm{g\,mol^{-1}}$) has a density of $1.08\ \mathrm{g\,mL^{-1}}$. The molality of the solution (in $\mathrm{mol\,kg^{-1}}$) is ______.
E3. NEET An aqueous solution is heated from $20\ ^{\circ}\mathrm{C}$ to $50\ ^{\circ}\mathrm{C}$ with no loss of water. Which pair of concentration terms both remain unchanged? (A) molarity and w/v % (B) molality and mole fraction (C) molarity and molality (D) mole fraction and v/v %
E4. JEE Main NEET $K_{H}$ values in water at $293\ \mathrm{K}$: He $145\ \mathrm{kbar}$, $\mathrm{H_{2}}$ $69.2\ \mathrm{kbar}$, $\mathrm{N_{2}}$ $76.5\ \mathrm{kbar}$, $\mathrm{O_{2}}$ $34.9\ \mathrm{kbar}$. At the same partial pressure, the correct order of solubility is (A) He $\gt$ $\mathrm{N_{2}}$ $\gt$ $\mathrm{H_{2}}$ $\gt$ $\mathrm{O_{2}}$ (B) $\mathrm{O_{2}}$ $\gt$ $\mathrm{H_{2}}$ $\gt$ $\mathrm{N_{2}}$ $\gt$ He (C) $\mathrm{O_{2}}$ $\gt$ $\mathrm{N_{2}}$ $\gt$ $\mathrm{H_{2}}$ $\gt$ He (D) He $\gt$ $\mathrm{H_{2}}$ $\gt$ $\mathrm{N_{2}}$ $\gt$ $\mathrm{O_{2}}$
E5. JEE Advanced A diver breathes compressed air ($78\%$ $\mathrm{N_{2}}$ by volume) at a depth where the total pressure is $4.0\ \mathrm{bar}$, until her blood is saturated with $\mathrm{N_{2}}$. She then rises rapidly to the surface ($1.0\ \mathrm{bar}$). Treat blood as water ($1.0\ \mathrm{L}\approx55.6\ \mathrm{mol}$) and use $K_{H}(\mathrm{N_{2}})=76.48\ \mathrm{kbar}$ throughout. Estimate the volume of $\mathrm{N_{2}}$ gas, measured at $310\ \mathrm{K}$ and $1.0\ \mathrm{bar}$, released from each litre of blood. ($R=0.083\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$.)
E6. Board NEET Assertion–reason. Assertion (A): Fish and other aquatic animals are more comfortable in cold water than in warm water. Reason (R): The value of $K_{H}$ for oxygen in water increases with temperature. (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.
Research question. How much does warming reduce the oxygen that a river or lake can hold, and how well does Henry's law predict the values that environmental agencies measure?
Why it matters. Most fish need at least about $5\ \mathrm{mg\,L^{-1}}$ of dissolved oxygen (DO). Water-quality standards in many countries, including India's criteria for bathing water, use this figure. Heat waves, warm water from power stations and slow summer flows all lower DO. Fish die-offs in rivers and lakes often follow.
What to do.
- Use Henry's law with $p_{\mathrm{O_{2}}}=0.21\ \mathrm{bar}$ to compute the DO (in $\mathrm{mg\,L^{-1}}$) of air-saturated water at $293\ \mathrm{K}$ and $303\ \mathrm{K}$ from the table ($K_{H}=34.86$ and $46.82\ \mathrm{kbar}$), as in Example 5.
- Estimate $K_{H}$ at $298\ \mathrm{K}$ by assuming it changes linearly between the two values, and compute the DO there too.
- Measured DO values for air-saturated fresh water at sea level are about $9.1$, $8.3$ and $7.6\ \mathrm{mg\,L^{-1}}$ at $20$, $25$ and $30\ ^{\circ}\mathrm{C}$. Tabulate your predictions beside them and find the percentage difference at each temperature.
- A river reservoir at $2000\ \mathrm{m}$ altitude has an air pressure of about $0.79\ \mathrm{bar}$. Scale your $303\ \mathrm{K}$ prediction to this pressure. Is the $5\ \mathrm{mg\,L^{-1}}$ threshold still met when the water is fully saturated with air?
Tools: Calculator or spreadsheet; Interactive 2 to check each $x$.
Hand in: A half-page note with your table, a sentence on how good Henry's law is here, and one practical step (shade trees, aeration, limits on warm discharges) that would protect fish.
Going further. Real water is never quite saturated: decaying organic matter uses up oxygen (biochemical oxygen demand, BOD). Find the published DO and BOD records for a river near you, or for a well-studied river such as the Thames or the Ganga. Compare the summer and winter values.
🧪 Self-check quiz
Q1. The molality of a solution is the number of moles of solute per
Q2. Which concentration term does not change when a solution is heated?
Q3. A solution contains $1\ \mathrm{mol}$ of glucose and $9\ \mathrm{mol}$ of water. The mole fraction of glucose is
Q4. Gas P has $K_{H}=2\ \mathrm{kbar}$ and gas Q has $K_{H}=80\ \mathrm{kbar}$ in water at the same temperature. At equal partial pressures,
Q5. "The bends" in divers who surface too quickly is caused by