Chemistry Solutions

Solutions Formula Sheet

All key Solutions formulas for JEE: concentration units, Raoult's law, colligative properties, osmotic pressure, and van't Hoff factor — quick revision for JEE Main & Advanced.

7 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know formula, constant, and high-yield fact from the Solutions chapter in one scannable sheet. Use it for last-minute revision before JEE Main and Advanced.

Concentration Units

QuantityFormulaNotes
Molarity (M)$M = \dfrac{n}{V_{\text{(L)}}}$mol per L of solution; temperature dependent
Molality (m)$m = \dfrac{n}{W_{\text{solvent (kg)}}}$mol per kg of solvent; temperature independent
Mole fraction ($\chi$)$\chi_A = \dfrac{n_A}{n_{\text{total}}}$Dimensionless; $\chi_{\text{solute}} + \chi_{\text{solvent}} = 1$
ppm$\text{ppm} = \dfrac{\text{mass solute}}{\text{mass solution}} \times 10^6$Trace concentrations
% (w/w)$\dfrac{\text{mass solute}}{\text{mass solution}} \times 100$Temperature independent
% (v/v)$\dfrac{\text{volume solute}}{\text{volume solution}} \times 100$Temperature dependent
% (w/v)$\dfrac{\text{mass solute (g)}}{\text{volume solution (mL)}} \times 100$Temperature dependent

Molarity from Mass

$$\boxed{M = \frac{W_{\text{solute}} \times 1000}{M_{\text{solute}} \times V_{\text{(mL)}}}}$$$$\text{Moles} = M \times V_{\text{(L)}} = \frac{M \times V_{\text{(mL)}}}{1000}$$

Molality from Mass

$$\boxed{m = \frac{W_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent (g)}}}}$$
High-Yield Memory Hook

M for Mix the Volume, m for Mass of solvent. Molarity uses total solution volume; molality uses solvent mass. Molality is the one used in colligative properties because it doesn’t change with temperature.

Interconversion Formulas

ConversionFormulaNotes
M $\to$ m$m = \dfrac{1000M}{1000d - M \cdot M_{\text{solute}}}$$d$ = density (g/mL)
m $\to$ M$M = \dfrac{1000\,m\,d}{1000 + m \cdot M_{\text{solute}}}$Density is the bridge
m $\to \chi_{\text{solute}}$$\chi_{\text{solute}} = \dfrac{m \cdot M_{\text{solvent}}}{1000 + m \cdot M_{\text{solvent}}}$$M_{\text{solvent}}$ in g/mol
m $\to \chi_{\text{solvent}}$$\chi_{\text{solvent}} = \dfrac{1000}{1000 + m \cdot M_{\text{solvent}}}$
M $\to \chi_{\text{solute}}$$\chi_{\text{solute}} = \dfrac{M \cdot M_{\text{solvent}}}{1000\,d}$For dilute aqueous ($d \approx 1$)
Normality $\leftrightarrow$ Molarity$N = n \times M$$n$ = n-factor (valency factor)

n-factor (valency factor): number of H⁺ for acids, OH⁻ for bases, or change in oxidation state for redox.

Raoult’s Law (Ideal Solutions)

$$\boxed{P_A = P_A^0 \cdot \chi_A}$$$$\boxed{P_{\text{total}} = P_A^0 \chi_A + P_B^0 \chi_B}$$

For a non-volatile solute ($P_B^0 = 0$):

$$\boxed{P_{\text{solution}} = P_{\text{solvent}}^0 \cdot \chi_{\text{solvent}} = P_{\text{solvent}}^0 (1 - \chi_{\text{solute}})}$$

Vapor-Phase Composition (Dalton’s Law)

$$\boxed{y_A = \frac{P_A}{P_{\text{total}}} = \frac{P_A^0 \chi_A}{P_A^0 \chi_A + P_B^0 \chi_B}}$$
Vapor Enrichment

The vapor is always richer in the more volatile component: if $P_A^0 > P_B^0$, then $y_A > \chi_A$. This is the basis of fractional distillation.

Ideal vs Non-Ideal Solutions

PropertyIdealPositive DeviationNegative Deviation
$\Delta H_{\text{mix}}$0$> 0$ (endothermic)$< 0$ (exothermic)
$\Delta V_{\text{mix}}$0$> 0$$< 0$
A–B vs A–A, B–BEqualWeakerStronger
$P_{\text{total}}$Obeys Raoult$> P_{\text{ideal}}$$< P_{\text{ideal}}$
ExampleBenzene + TolueneEthanol + WaterChloroform + Acetone
AzeotropeMinimum boilingMaximum boiling

Must-remember examples

  • Ideal: Benzene + Toluene; n-Hexane + n-Heptane; CCl₄ + SiCl₄
  • Positive deviation: Ethanol + Water; Acetone + CS₂; CCl₄ + CHCl₃
  • Negative deviation: Chloroform + Acetone (new H-bond); Water + HNO₃; Water + HCl

Azeotropes (cannot be separated by fractional distillation)

TypeDeviationExample (composition, BP)
Minimum boilingPositive (max VP)Ethanol (95.6%) + Water → 78.2°C
Maximum boilingNegative (min VP)HNO₃ (68%) + Water → 120.5°C

Colligative Properties

The four colligative properties depend only on the number of solute particles, not their nature. Mnemonic: ROBE — RLVP, Osmotic pressure, Boiling-point elevation, freezing-point dEpression.

1. Relative Lowering of Vapor Pressure (RLVP)

$$\boxed{\frac{P^0 - P_s}{P^0} = \chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}}$$

For dilute solutions ($n_{\text{solute}} \ll n_{\text{solvent}}$):

$$\frac{P^0 - P_s}{P^0} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{w/M}{W/M_{\text{solvent}}}$$

For dilute aqueous solutions ($M_{\text{water}} = 18$):

$$\frac{\Delta P}{P^0} = \frac{w \times 18}{M \times W}$$

2. Elevation of Boiling Point

$$\boxed{\Delta T_b = K_b \, m = K_b \cdot \frac{w \times 1000}{M \times W}}$$

3. Depression of Freezing Point

$$\boxed{\Delta T_f = K_f \, m = K_f \cdot \frac{w \times 1000}{M \times W}}$$

where $w$ = mass of solute (g), $M$ = molar mass of solute, $W$ = mass of solvent (g).

Theoretical Relations for $K_b$ and $K_f$

$$K_b = \frac{R \, M_{\text{solvent}} \, (T_b^0)^2}{1000 \, \Delta H_{\text{vap}}} \qquad K_f = \frac{R \, M_{\text{solvent}} \, (T_f^0)^2}{1000 \, \Delta H_{\text{fusion}}}$$

Both depend on solvent properties only. Since $\Delta H_{\text{fusion}} < \Delta H_{\text{vap}}$, usually $K_f > K_b$.

Ratio of the Two Constants (same molality)

$$\boxed{\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}}$$

For water: $\dfrac{\Delta T_b}{\Delta T_f} = \dfrac{0.52}{1.86} \approx \dfrac{1}{3.58}$ — FP depression is ~3.5× the BP elevation.

Molal Constants (memorize)

Solvent$K_b$ (K·kg/mol)Normal BP (°C)$K_f$ (K·kg/mol)Normal FP (°C)
Water0.521001.860
Benzene2.5380.15.125.5
Chloroform3.6361.24.68−63.5
Ethanol1.2278.4
CCl₄5.0376.7
Acetic acid3.9016.6
Camphor37.7179.8
Constants to Lock In

Water: $K_b = 0.52$, $K_f = 1.86$. Benzene: $K_b = 2.53$, $K_f = 5.12$. Camphor $K_f = 37.7$ (highest) — used for molar-mass determination by Rast method.

4. Osmotic Pressure

$$\boxed{\pi = CRT = \frac{n}{V}RT = \frac{w}{M\,V}RT}$$

where $C$ = molarity (mol/L), $R$ = 0.0821 L·atm/(mol·K) or 8.314 J/(mol·K), $T$ in K, $V$ in L.

Isotonic condition: $\;\pi_1 = \pi_2 \implies C_1 = C_2$ (same T)

Reverse osmosis: $\;P_{\text{applied}} > \pi$

TonicityConcentration vs cellWater flowCell response
HypotonicLower (e.g. < 0.9% NaCl)Into cellSwells, may burst (hemolysis)
IsotonicEqual (0.9% NaCl)No net flowNormal
HypertonicHigher (e.g. > 0.9% NaCl)Out of cellShrinks (crenation)
Osmotic Pressure Pitfalls

$\pi$ uses molarity (C), not molality. Temperature must be in Kelvin. Pick the right $R$: use 0.0821 with atm, 8.314 with Pa. Water flows from dilute → concentrated (low $\pi$ → high $\pi$).

Molar Mass Determination

FromFormula
Boiling-point elevation$M = \dfrac{K_b \, w \times 1000}{\Delta T_b \times W}$
Freezing-point depression$M = \dfrac{K_f \, w \times 1000}{\Delta T_f \times W}$
Osmotic pressure$M = \dfrac{w\,R\,T}{\pi\,V}$

Osmotic pressure is best for macromolecules (polymers, proteins) — large, easily measured $\pi$ at room temperature.

van’t Hoff Factor (i)

$$\boxed{i = \frac{\text{Actual no. of particles in solution}}{\text{No. of formula units dissolved}} = \frac{\text{Observed colligative property}}{\text{Calculated (assuming } i = 1)}}$$

Modified Colligative Properties

$$\frac{P^0 - P_s}{P^0} = i\,\chi_{\text{solute}} \qquad \Delta T_b = i\,K_b\,m \qquad \Delta T_f = i\,K_f\,m \qquad \pi = i\,CRT$$

Dissociation (i > 1)

$$\boxed{i = 1 + \alpha(n - 1)} \qquad \boxed{\alpha = \frac{i - 1}{n - 1}}$$

where $n$ = number of ions on complete dissociation, $\alpha$ = degree of dissociation.

CompoundDissociationTheoretical i
NaCl, KBr$\to$ 2 ions2
CaCl₂, Na₂SO₄, K₂SO₄$\to$ 3 ions3
AlCl₃, K₃[Fe(CN)₆]$\to$ 4 ions4
K₄[Fe(CN)₆], Al₂(SO₄)₃$\to$ 5 ions5

Association (i < 1)

$$\boxed{i = 1 - \alpha\left(1 - \frac{1}{n}\right)} \qquad \boxed{\alpha = \frac{(1-i)\,n}{n-1}}$$

For dimerization ($n = 2$):

$$\boxed{i = 1 - \frac{\alpha}{2}} \qquad \boxed{\alpha = 2(1 - i)}$$

For trimerization ($n = 3$): $\;i = 1 - \dfrac{2\alpha}{3}$

Abnormal Molar Mass

$$\boxed{i = \frac{M_{\text{theoretical}}}{M_{\text{observed}}}} \qquad M_{\text{observed}} = \frac{M_{\text{theoretical}}}{i}$$
Processi$M_{\text{observed}}$
Dissociation$> 1$$< M_{\text{theoretical}}$
Association$< 1$$> M_{\text{theoretical}}$
Neither$= 1$$= M_{\text{theoretical}}$
i Decision Tree

$i > 1 \Rightarrow$ dissociation (electrolyte). $i = 1 \Rightarrow$ non-electrolyte (glucose, urea, sucrose). $i < 1 \Rightarrow$ association (e.g. acetic acid / benzoic acid in benzene, $i \approx 0.5$). Don’t mix up the dissociation and association formulas.

Real-World i Values

SubstanceBehaviori
Glucose / urea / sucrose (water)Non-electrolyte1
NaCl (dilute)Strong electrolyte~1.9–2.0
NaCl (1 M)Strong electrolyte~1.7 (ion-pairing)
CaCl₂Strong electrolyte~2.5–2.7
CH₃COOH (water)Weak acid1.01–1.10
CH₃COOH / benzoic acid (benzene)Associated (dimer)~0.5

Henry’s Law (Gas in Liquid)

$$\boxed{P_{\text{gas}} = K_H \cdot \chi_{\text{gas}}}$$

Higher $K_H$ $\to$ lower solubility. Applications: carbonated drinks (CO₂), scuba diving (the bends / N₂ narcosis), O₂ solubility in blood. Example: $K_H$ for CO₂ in water at 25°C $= 1.67 \times 10^8$ torr.

Key Constants & Reference Facts

ItemValue
$R$0.0821 L·atm/(mol·K) = 8.314 J/(mol·K)
Water: $M$, density18 g/mol, 1 g/mL (≈ 55.56 mol per L)
Normal saline0.9% NaCl — isotonic with blood ($\pi \approx 7.8$ atm)
Isotonic glucose~5% (w/v)
Seawater~3.5% salt → $\pi \approx 30$ atm
Common molar massesNaCl 58.5, H₂SO₄ 98, NaOH 40, glucose 180, urea 60, sucrose 342
JEE Exam Strategy

Highest-weightage items across this chapter: M↔m interconversion, RLVP with non-volatile solute, molar mass from $\Delta T_f$, $\pi = iCRT$ for electrolytes, and computing $i$ then $\alpha$. Always check: molality (not molarity) for $\Delta T_b/\Delta T_f$, molarity (not molality) for $\pi$, the ×1000 factor, $\Delta T_f = T_f^0 - T_f$, and the van’t Hoff factor for any electrolyte.