Solutions • Topic 1 of 3

Concentration & Solubility

A solution is a homogeneous mixture of two or more components whose composition can be varied within limits. The component present in the largest amount is the solvent; the other components are solutes. Depending on the physical state of solvent and solute we get nine types of solutions — gas in gas (air), gas in liquid (soda water), solid in liquid (salt in water), solid in solid (alloys like brass) and so on. In Class 12 we focus mostly on liquid solutions, especially solids and gases dissolved in liquids.

To do quantitative work we must express concentration precisely. The common terms are:

  • Mass percentage $(w/w)$: mass of component per 100 g of solution, i.e. $\text{mass \%}=\frac{\text{mass of component}}{\text{mass of solution}}\times100$.
  • Mole fraction $(x)$: for a component, $x_A=\frac{n_A}{n_A+n_B}$, and $x_A+x_B=1$. Mole fraction is dimensionless and independent of temperature.
  • Molarity $(M)$: moles of solute per litre of solution, $M=\frac{n_{solute}}{V_{solution}(\text{L})}$. It is temperature-dependent because volume changes with temperature.
  • Molality $(m)$: moles of solute per kilogram of solvent, $m=\frac{n_{solute}}{w_{solvent}(\text{kg})}$. Being mass-based it is temperature-independent, so it is preferred in colligative-property calculations.
  • Parts per million $(\text{ppm})$: used for trace amounts, $\text{ppm}=\frac{\text{mass of component}}{\text{mass of solution}}\times10^6$.

Solubility is the maximum amount of solute that dissolves in a given amount of solvent at a specified temperature to form a saturated solution. For solids in liquids, solubility usually increases with temperature for endothermic dissolution (e.g. $\text{KNO}_3$) and may decrease for exothermic dissolution. Pressure has almost no effect on solids.

For gases in liquids, solubility increases with pressure and decreases with rising temperature (which is why warm soda goes flat). The quantitative law is Henry's law: the partial pressure of a gas over a solution is proportional to its mole fraction in solution, $p=K_H\,x$, where $K_H$ is Henry's constant. A higher $K_H$ means lower solubility of the gas. Henry's law explains the bends in deep-sea divers, the use of helium-oxygen mixtures, and oxygen transport in blood.

Concentration terms at a glance
TermDefinitionFormulaTemp-dependent?
Mass %mass per 100 g solution(w/w)×100No
Mole fraction (x)moles of component / total molesn_A/(n_A+n_B)No
Molarity (M)moles solute / L solutionn/V(L)Yes
Molality (m)moles solute / kg solventn/w(kg)No
ppmparts per million by mass(w/w)×10^6No
Solved Examples
1
Worked Example
Calculate the mole fraction of ethylene glycol $(\text{C}_2\text{H}_6\text{O}_2)$ in a solution containing 20% of it by mass.
Solution
  1. Take 100 g of solution: glycol = 20 g, water = 80 g.
  2. Moles of glycol $=\frac{20}{62}=0.322$ mol (M of glycol = 62).
  3. Moles of water $=\frac{80}{18}=4.444$ mol.
  4. $x_{glycol}=\frac{0.322}{0.322+4.444}=\frac{0.322}{4.766}=0.0676$.

Answer: $x_{glycol}\approx0.068$.

2
Worked Example
A solution is prepared by dissolving 5 g of $\text{NaOH}$ in water to make 450 mL of solution. Find its molarity. (M of NaOH = 40)
Solution
  1. Moles of NaOH $=\frac{5}{40}=0.125$ mol.
  2. Volume $=450\ \text{mL}=0.450\ \text{L}$.
  3. $M=\frac{0.125}{0.450}=0.278\ \text{mol L}^{-1}$.

Answer: $M\approx0.28\ \text{M}$.

3
Worked Example
Calculate the molality of a solution containing 2.5 g of urea $(\text{M}=60)$ dissolved in 100 g of water.
Solution
  1. Moles of urea $=\frac{2.5}{60}=0.0417$ mol.
  2. Mass of solvent $=100\ \text{g}=0.100\ \text{kg}$.
  3. $m=\frac{0.0417}{0.100}=0.417\ \text{mol kg}^{-1}$.

Answer: $m\approx0.42\ \text{m}$.

4
Worked Example
The density of a 3 M solution of $\text{NaCl}$ (M = 58.5) is $1.25\ \text{g mL}^{-1}$. Calculate its molality.
Solution
  1. 3 M means 3 mol NaCl in 1 L solution; mass of NaCl $=3\times58.5=175.5$ g.
  2. Mass of 1 L solution $=1000\times1.25=1250$ g.
  3. Mass of water $=1250-175.5=1074.5\ \text{g}=1.0745\ \text{kg}$.
  4. $m=\frac{3}{1.0745}=2.79\ \text{mol kg}^{-1}$.

Answer: $m\approx2.79\ \text{m}$.

5
Worked Example
If $\text{N}_2$ gas is bubbled through water at 293 K, how many millimoles of $\text{N}_2$ dissolve in 1 litre of water at a partial pressure of 0.987 bar? $K_H$ for $\text{N}_2$ at 293 K is $76.48\ \text{kbar}$.
Solution
  1. Henry's law: $x_{N_2}=\frac{p}{K_H}=\frac{0.987}{76480}=1.29\times10^{-5}$.
  2. 1 L water = 1000 g $\Rightarrow n_{water}=\frac{1000}{18}=55.5$ mol; $n_{N_2}\ll n_{water}$, so $x_{N_2}\approx\frac{n_{N_2}}{55.5}$.
  3. $n_{N_2}=1.29\times10^{-5}\times55.5=7.16\times10^{-4}\ \text{mol}$.
  4. In millimoles $=7.16\times10^{-4}\times1000=0.716\ \text{mmol}$.

Answer: $\approx0.716\ \text{mmol of N}_2$.

6
Worked Example
Henry's constant for $\text{CO}_2$ in water is $1.67\times10^8\ \text{Pa}$ at 298 K. Calculate the mole fraction of $\text{CO}_2$ in soda water under a $\text{CO}_2$ pressure of $2.5$ atm. (1 atm = $1.013\times10^5$ Pa)
Solution
  1. Convert pressure: $p=2.5\times1.013\times10^5=2.53\times10^5\ \text{Pa}$.
  2. Henry's law: $x_{CO_2}=\frac{p}{K_H}=\frac{2.53\times10^5}{1.67\times10^8}$.
  3. $x_{CO_2}=1.52\times10^{-3}$.

Answer: $x_{CO_2}\approx1.52\times10^{-3}$.

Key Points

  • A solution is a homogeneous mixture; the solvent is the major component, the solute(s) the minor — nine types exist by physical state.
  • Mole fraction $x$ and molality $m$ are temperature-independent (mass-based); molarity $M$ is temperature-dependent because volume varies.
  • Molality $m=\frac{n_{solute}}{w_{solvent}(\text{kg})}$ is preferred for colligative properties; ppm is used for trace concentrations.
  • Solid solubility usually rises with temperature (endothermic dissolution) and is nearly pressure-independent.
  • Henry's law $p=K_H x$: gas solubility rises with pressure and falls with temperature; a larger $K_H$ means a less soluble gas.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Which concentration term is independent of temperature?
Explanation: Molality is defined per kilogram of solvent (mass), which does not change with temperature, unlike volume-based molarity.
Q2.According to Henry's law, the solubility of a gas in a liquid increases when:
Explanation: Since $p=K_H x$, a higher partial pressure gives a higher mole fraction (solubility). Higher temperature or higher $K_H$ both lower solubility.
Q3.A larger value of Henry's constant $K_H$ for a gas indicates:
Explanation: From $x=p/K_H$, a larger $K_H$ gives a smaller mole fraction at the same pressure, i.e. the gas is less soluble.
Q4.The mole fraction of solute in a dilute aqueous solution containing 0.1 mol solute in 900 g water is approximately:
Explanation: $n_{water}=900/18=50$ mol; $x_{solute}=0.1/(0.1+50)\approx0.002$.
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