Why Concentration Needs a Unit
A solution is a homogeneous mixture of two or more components whose composition and properties are uniform throughout. In a binary solution the component present in the largest quantity is the solvent, and the remaining component dissolved in it is the solute. The same pair of substances, however, can be combined in countless proportions, and the behaviour of the mixture changes sharply with that proportion.
NCERT illustrates this with fluoride ions in water: 1 part per million prevents tooth decay, 1.5 ppm causes mottling of teeth, and high concentrations are poisonous. Qualitative words such as "dilute" or "concentrated" cannot capture such fine distinctions, so chemistry insists on a quantitative description. The choice of which unit to use depends on what the number must remain reliable against — most importantly, whether it must survive a change in temperature.
Percentage and ppm Units
The most intuitive descriptions of concentration are the percentage units, each defined as the amount of one component relative to the whole solution, scaled to 100. NCERT defines three percentage variants plus parts per million for trace amounts.
| Unit | Definition | Worked meaning |
|---|---|---|
| Mass % (w/w) | $\dfrac{\text{Mass of component}}{\text{Total mass of solution}}\times 100$ | 10% glucose by mass = 10 g $\ce{C6H12O6}$ in 90 g water (100 g solution). |
| Volume % (v/v) | $\dfrac{\text{Volume of component}}{\text{Total volume of solution}}\times 100$ | 10% ethanol = 10 mL ethanol made up to 100 mL solution. |
| Mass by volume % (w/V) | $\dfrac{\text{Mass of solute}}{100\ \text{mL of solution}}$ | Common in medicine and pharmacy; mass of solute per 100 mL solution. |
| ppm | $\dfrac{\text{Parts of component}}{\text{Total parts of solution}}\times 10^{6}$ | Sea water with $5.8\ \text{g}$ of $\ce{O2}$ per $10^{6}$ g of water = 5.8 ppm. |
Mass percentage is favoured in industrial chemistry — commercial bleaching solution is quoted as 3.62 mass percent of $\ce{NaOCl}$ in water. Volume percentage suits liquid-in-liquid mixtures such as the 35% (v/v) ethylene glycol antifreeze used in car radiators. Parts per million handles trace solutes such as dissolved gases and pollutants, and like percentage it may be expressed mass-to-mass, volume-to-volume or mass-to-volume.
Mole Fraction
Mole fraction, symbol $x$ with a subscript naming the component, expresses concentration in terms of the relative number of particles rather than mass or volume. It is the workhorse unit for relating vapour pressure and other physical properties to composition.
$$ x_i = \frac{n_i}{n_1 + n_2 + \cdots + n_i} = \frac{n_i}{\sum n_i} $$For a binary mixture of components A and B with $n_A$ and $n_B$ moles, the mole fraction of A is $x_A = \dfrac{n_A}{n_A + n_B}$. A defining feature follows directly from the formula: the mole fractions of all components sum to unity.
$$ x_1 + x_2 + \cdots + x_i = 1 $$Calculate the mole fraction of ethylene glycol ($\ce{C2H6O2}$) in a solution containing 20% of $\ce{C2H6O2}$ by mass. (NCERT Example 1.1)
Take 100 g of solution → 20 g of $\ce{C2H6O2}$ and 80 g of $\ce{H2O}$.
Molar mass of $\ce{C2H6O2} = 12\times2 + 1\times6 + 16\times2 = 62\ \text{g mol}^{-1}$.
$n_{\text{glycol}} = \dfrac{20}{62} = 0.322\ \text{mol}; \quad n_{\text{water}} = \dfrac{80}{18} = 4.444\ \text{mol}$
$x_{\text{glycol}} = \dfrac{0.322}{0.322 + 4.444} = \mathbf{0.068}$
$x_{\text{water}} = \dfrac{4.444}{0.322 + 4.444} = 0.932$, which also equals $1 - 0.068 = 0.932$.
Molarity
Molarity (M) is the number of moles of solute dissolved in one litre — equivalently one cubic decimetre — of solution. It is the most widely used laboratory unit because solution volume is easy to measure with a flask or pipette.
$$ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}} $$A 0.25 mol L$^{-1}$ (0.25 M) solution of $\ce{NaOH}$ contains 0.25 mol of $\ce{NaOH}$ in one litre of solution. Because the denominator is a volume, molarity carries a hidden temperature sensitivity that becomes important later in this article.
Calculate the molarity of a solution containing 5 g of $\ce{NaOH}$ in 450 mL of solution. (NCERT Example 1.2)
Molar mass of $\ce{NaOH} = 40\ \text{g mol}^{-1}$, so $n_{\ce{NaOH}} = \dfrac{5}{40} = 0.125\ \text{mol}$.
Volume of solution $= \dfrac{450}{1000} = 0.450\ \text{L}$.
$\text{Molarity} = \dfrac{0.125\ \text{mol}}{0.450\ \text{L}} = \mathbf{0.278\ M} = 0.278\ \text{mol dm}^{-3}$
Concentration sets the stage for solubility limits. See how much solute a solvent can actually hold in Solubility of Solids and Gases.
Molality
Molality (m) is the number of moles of solute per kilogram of solvent. Crucially, its denominator is a mass of solvent, not a volume of solution and not the total mass — a distinction that NEET examiners probe directly.
$$ \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} $$A 1.00 mol kg$^{-1}$ (1.00 m) solution of $\ce{KCl}$ contains 1 mol — that is 74.5 g — of $\ce{KCl}$ dissolved in 1 kg of water. Because both numerator and denominator are mass-based quantities, molality is completely insensitive to temperature, which makes it the unit of choice for colligative-property calculations.
Calculate the molality of a solution of 2.5 g of ethanoic acid ($\ce{CH3COOH}$) in 75 g of benzene. (NCERT Example 1.3)
Molar mass of $\ce{CH3COOH} = 12\times2 + 1\times4 + 16\times2 = 60\ \text{g mol}^{-1}$.
$n_{\text{acid}} = \dfrac{2.5}{60} = 0.0417\ \text{mol}$.
Mass of benzene (solvent) $= \dfrac{75}{1000} = 0.075\ \text{kg}$.
$\text{Molality} = \dfrac{0.0417\ \text{mol}}{0.075\ \text{kg}} = \mathbf{0.556\ mol\ kg^{-1}}$
Molarity is temperature dependent — molality and mole fraction are not
The single most-asked discriminator in this topic. Volume expands when heated, so any unit built on solution volume changes with temperature. Mass and mole counts do not change with temperature, so any unit built only from masses or moles stays fixed.
Temperature dependent: molarity (M), volume %. Temperature independent: mass %, ppm, mole fraction, molality (m).
Master Table of Concentration Units
The following master table consolidates every unit in NCERT §1.2 against its formula, its unit and — the column NEET cares about most — its behaviour with temperature.
| Term | Formula | Unit | Temperature dependence |
|---|---|---|---|
| Mass percentage (w/w) | $\dfrac{\text{mass of component}}{\text{total mass}}\times 100$ | % | Independent |
| Volume percentage (v/v) | $\dfrac{\text{volume of component}}{\text{total volume}}\times 100$ | % | Dependent |
| Mass by volume (w/V) | $\dfrac{\text{mass of solute}}{100\ \text{mL solution}}$ | g / 100 mL | Dependent |
| Parts per million (ppm) | $\dfrac{\text{parts of component}}{\text{total parts}}\times 10^{6}$ | ppm | Independent (mass basis) |
| Mole fraction ($x$) | $\dfrac{n_i}{\sum n_i}$ | dimensionless | Independent |
| Molarity (M) | $\dfrac{\text{moles of solute}}{\text{volume of solution (L)}}$ | mol L$^{-1}$ (mol dm$^{-3}$) | Dependent |
| Molality (m) | $\dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}}$ | mol kg$^{-1}$ | Independent |
Temperature Dependence
NCERT states the rule explicitly: mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity is a function of temperature. The reasoning is purely physical and worth internalising rather than memorising.
The same five solute particles (fixed moles, fixed mass) occupy a larger volume at higher temperature. Since molarity divides moles by volume, the value drops while molality — which divides by solvent mass — stays unchanged.
This is exactly why colligative properties — elevation of boiling point, depression of freezing point and osmotic effects, all of which are measured at temperatures that differ from the preparation temperature — are framed in molality or mole fraction rather than molarity. A molarity value quoted at 25 °C would no longer be accurate at the boiling point of the solution.
Interconversion Using Density
Volume-based units (molarity) and mass-based units (molality, mole fraction) cannot be converted into one another by counting moles alone — the link between volume and mass is the density of the solution. The standard NEET strategy is to assume a convenient fixed amount of solution, compute every mass and mole quantity, and then read off whichever unit is asked.
Density converts a known solution volume into a known solution mass, unlocking conversion in either direction between molarity and the mass-based units.
Calculate (a) molality, (b) molarity and (c) mole fraction of $\ce{KI}$ if the density of a 20% (mass/mass) aqueous $\ce{KI}$ solution is 1.202 g mL$^{-1}$. (NCERT Example 1.5)
Set-up. Take 100 g of solution. Then it contains 20 g of $\ce{KI}$ and 80 g of $\ce{H2O}$. Molar mass of $\ce{KI} = 39 + 127 = 166\ \text{g mol}^{-1}$.
$n_{\ce{KI}} = \dfrac{20}{166} = 0.120\ \text{mol}; \quad n_{\ce{H2O}} = \dfrac{80}{18} = 4.444\ \text{mol}$.
(a) Molality — mass of solvent $= 80\ \text{g} = 0.080\ \text{kg}$:
$m = \dfrac{0.120\ \text{mol}}{0.080\ \text{kg}} = \mathbf{1.51\ mol\ kg^{-1}}$
(b) Molarity — use density to find the volume of 100 g of solution: $V = \dfrac{\text{mass}}{\text{density}} = \dfrac{100\ \text{g}}{1.202\ \text{g mL}^{-1}} = 83.2\ \text{mL} = 0.0832\ \text{L}$:
$M = \dfrac{0.120\ \text{mol}}{0.0832\ \text{L}} = \mathbf{1.44\ mol\ L^{-1}}$
(c) Mole fraction of $\ce{KI}$:
$x_{\ce{KI}} = \dfrac{0.120}{0.120 + 4.444} = \mathbf{0.0263}$
Note how the density was needed only for molarity — the mass-based units (a) and (c) never used it, confirming their independence from volume.
A useful generalisation drops out of this method. For a solution of molarity $M$, molar mass of solute $M_{\!B}$ and density $d$ (in g mL$^{-1}$), the molality is obtained by isolating the solvent mass from one litre of solution:
$$ m = \frac{1000\,M}{1000\,d - M\,M_{\!B}} $$Here $1000\,d$ is the mass of one litre of solution in grams and $M\,M_{\!B}$ is the mass of dissolved solute, so the difference is the solvent mass. For dilute solutions $M\,M_{\!B}$ is small and $d \approx 1$, which is why molarity and molality of dilute aqueous solutions are nearly equal — a fact several NEET items exploit by treating molarity as molality.
Lock these before the exam
- Molarity (M) = moles solute / litre of solution; unit mol L$^{-1}$; temperature dependent.
- Molality (m) = moles solute / kg of solvent; unit mol kg$^{-1}$; temperature independent.
- Mole fraction = $n_i / \sum n_i$; dimensionless; all components sum to 1; temperature independent.
- Mass % and ppm are mass-based and temperature independent; volume % and mass-by-volume are volume-based and temperature dependent.
- Use density to bridge molarity and molality; assume a fixed 100 g of solution for percentage problems.
- Colligative property questions are framed in molality / mole fraction, never molarity.