Concentration of Solutions

Concentration is the answer to a single question: how much solute is dissolved in how much solvent (or solution)? But that question has many sensible answers, depending on whether you're measuring mass, volume, moles, equivalents, or particle count. You'll meet seven different concentration units in this section. Each one is the same fact viewed through a different lens:

• % by weight (w/w) — mass-based
• % by volume (v/v) — volume-based
• Mole fraction (X) — pure ratios of particles, no units
• Molarity (M) — moles per litre of solution
• Normality (N) — equivalents per litre of solution
• Molality (m) — moles per kilogram of solvent
• PPM and PPB — parts-per-million and parts-per-billion, for very dilute solutions

These are all interchangeable — given enough information, you can convert between any two. The skill is knowing which unit to use for which problem, and how to convert between them quickly.

Three types of percentage concentration:

a) % by mass (w/w) — mass of solute per 100 g of solution:
$\% \text{ (w/w)} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

Example: 20% (w/w) NaCl → 20 g NaCl in 100 g solution (i.e., 20 g NaCl + 80 g $\ce{H2O}$)

b) % by volume (w/v) — mass of solute per 100 mL of solution:
$\% \text{ (w/v)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$

To prepare 20% (w/v) ethanol: put 20 g ethanol in a graduated cylinder and add water until total volume = 100 mL. You cannot simply mix 20 g ethanol + 80 mL water — volume is not an additive property.

c) % strength (v/v) — volume of solute per 100 mL of solution:
$\% \text{ (v/v)} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100$

Key fact: mass is additive, but volume is generally not (except for ideal solutions).

Molarity is the number of moles of solute present in 1 litre of solution.

$M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$

SI unit: mol L$^{-1}$ or Molar (abbreviated M)

Examples:

• 1 M HCl means 36.5 g HCl dissolved in enough water to make 1 L of solution
• 0.5 M NaOH means 20 g NaOH in 1 L solution

Key derived formulas:

• Moles of solute = Molarity × Volume (L)
• Millimoles of solute = Molarity × Volume (mL)

⚠️ Molarity is temperature-dependent — volume changes with temperature, so molarity of the same solution changes when temperature changes.

Dilution formula: When you dilute a solution, moles of solute are conserved:
$M_1 V_1 = M_2 V_2$

Example: To prepare 100 mL of 0.04 M $\ce{K2Cr2O7}$ from 0.2 M stock:
$V_1 = \frac{0.04 \times 100}{0.2} = 20$ mL of stock solution needed.

Two equivalent rearrangements of the molarity equation are worth memorising as separate facts:

$M \times V(\text{L}) = \text{number of moles}$

$M \times V(\text{mL}) = \text{number of millimoles}$

The second form is especially useful in titration problems, where volumes are naturally given in mL.

Caveat — molarity is temperature dependent. The volume of a solution expands when temperature rises, so the same mass of solute produces a lower molarity at higher temperatures. For colligative-property problems where temperature changes, switch to molality (which is mass-based and therefore temperature-independent).

Dilution is the process of adding more solvent to a solution to lower its concentration. The key principle: adding pure solvent doesn't change the number of moles of solute — only the total volume of solution increases.

• Before dilution: $n_{\text{solute}} = M_1 V_1$
• After dilution: $n_{\text{solute}} = M_2 V_2$

Since moles are conserved:

$\boxed{M_1 V_1 = M_2 V_2}$

One formula, every dilution problem. Plug in any three values and solve for the fourth.

Important caveat — this formula applies only to dilution (adding pure solvent to a single solution). When you mix two solutions of different concentrations, the formula becomes:

$M_1 V_1 + M_2 V_2 = M_{\text{final}} (V_1 + V_2)$

A classic JEE 2013 problem asked exactly this case: mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl gives $M = (0.5 \times 750 + 2 \times 250)/1000 = 0.875$ M. Don't apply $M_1 V_1 = M_2 V_2$ here — that's a different scenario.

Molality is the number of moles of solute per kilogram of solvent (not solution):

$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$

SI unit: mol kg$^{-1}$

✅ Molality is temperature-independent — it is based on mass, which does not change with temperature. This makes molality preferred for colligative property calculations.

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Mole Fraction is the ratio of moles of one component to the total moles of all components:

$X_A = \frac{n_A}{n_A + n_B} \quad \text{and} \quad X_B = \frac{n_B}{n_A + n_B}$

Important: $X_A + X_B = 1$ (mole fractions of all components always add up to 1)

Mole fraction ranges from 0 to 1 and has no units.

Example: A mixture has 3 mol A and 9 mol B:

• $X_A = \frac{3}{12} = 0.25$ (or 25%)
• $X_B = \frac{9}{12} = 0.75$ (or 75%)

Normality is the number of gram-equivalents (equivalents) of solute per litre of solution:

$N = \frac{\text{number of equivalents}}{\text{volume of solution (L)}} = M \times n\text{-factor}$

where the n-factor depends on the reaction:

• For acids: n-factor = number of replaceable $\ce{H^+}$ ions
• For bases: n-factor = number of replaceable $\ce{OH^-}$ ions
• For oxidising/reducing agents: n-factor = change in oxidation state per molecule

Examples: 0.2 M HCl → N = 0.2 × 1 = 0.2 N; 0.5 M $\ce{H2SO4}$ → N = 0.5 × 2 = 1 N

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PPM (Parts per Million) — used for very dilute solutions (pollutants, trace elements):
$\text{PPM} = \frac{\text{parts of solute}}{\text{parts of solution}} \times 10^6 = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$

PPB (Parts per Billion):
$\text{PPB} = \frac{\text{parts of solute}}{\text{parts of solution}} \times 10^9$