Solutions form a fundamental topic in JEE Chemistry and understanding them is essential for scoring well. This comprehensive guide covers the definitions, types, concentration terms, colligative properties, solubility, and related important formulas and concepts necessary for JEE aspirants.
A solution is a homogeneous mixture of two or more substances. It consists of:
Example: In a sugar solution, sugar is the solute and water is the solvent.
Solutions can be classified based on the physical state of solvent and solute:
Based on solubility:
Concentration indicates how much solute is present in a given amount of solution or solvent. Important terms are:
Molarity is defined as the number of moles of solute dissolved per liter of solution.
M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
Molality is the number of moles of solute dissolved per kilogram of solvent.
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
Normality is the number of equivalents of solute per liter of solution.
N = \frac{\text{equivalents of solute}}{\text{volume of solution in liters}}
Mole fraction of a component is the ratio of moles of that component to total moles in solution.
X_A = \frac{n_A}{\sum n_i}
Mass percentage of solute in solution is given by:
\% = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100
Volume percentage is used when both solute and solvent are liquids:
\% = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100
Solubility is the maximum amount of solute that can dissolve in a specific amount of solvent at a given temperature to form a saturated solution.
The solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
S = k_H \times P
Where,
S = solubility of gask_H = Henry’s constantP = partial pressure of the gasColligative properties depend on the number of solute particles, not their identity. These are very important for JEE.
Adding a non-volatile solute lowers the vapor pressure of the solvent.
P_{\text{solution}} = X_{\text{solvent}} \times P^0_{\text{solvent}}
Where, \( P^0_{\text{solvent}} \) is vapor pressure of pure solvent, \( X_{\text{solvent}} \) is mole fraction.
Boiling point of solution is higher than pure solvent.
\Delta T_b = K_b \times m \times i
Where,
Freezing point of solution is lower than pure solvent.
\Delta T_f = K_f \times m \times i
Where,
Osmotic pressure is the pressure required to stop the flow of solvent into solution through a semipermeable membrane.
\Pi = MRTi
Where,
Raoult’s law states that the partial vapor pressure of each volatile component in an ideal solution is directly proportional to its mole fraction.
P_A = X_A P_A^0
For an ideal solution,
P_{\text{total}} = P_A + P_B = X_A P_A^0 + X_B P_B^0
Deviations from Raoult’s law indicate non-ideal solutions showing positive or negative deviation.
Solubility product constant \( K_{sp} \) defines equilibrium between a solid and its ions in a saturated solution.
For a salt \( A_mB_n \) dissociating as:
A_mB_n (s) \leftrightarrow m A^{n+} + n B^{m-}
The solubility product is:
K_{sp} = [A^{n+}]^m [B^{m-}]^n
Knowledge of \( K_{sp} \) helps solve problems related to precipitation, solubility, and ionic equilibria.
The presence of a common ion decreases the solubility of a salt due to Le Chatelier's principle.
For example, solubility of \( \mathrm{AgCl} \) decreases if \( \mathrm{Cl}^- \) ions are added.
| Concept | Formula | Notes |
|---|---|---|
| Molarity (M) | \( M = \frac{\text{moles solute}}{\text{liters solution}} \) | Used when volume of solution known |
| Molality (m) | \( m = \frac{\text{moles solute}}{\text{kg solvent}} \) | Used when mass of solvent known, temperature independent |
| Normality (N) | \( N = \frac{\text{equivalents solute}}{\text{liters solution}} \) | Depends on reaction type |
| Mole Fraction (X) | \( X = \frac{\text{moles component}}{\text{total moles}} \) | Used in vapor pressure calculations |
| Boiling Point Elevation | \( \Delta T_b = K_b m i \) | Colligative property |
| Freezing Point Depression | \( \Delta T_f = K_f m i \) | Colligative property |
| Osmotic Pressure | \( \Pi = MRTi \) | Colligative property |
| Raoult’s Law | \( P_A = X_A P_A^0 \) | Ideal solutions vapor pressure |
| Henry’s Law | \( S = k_H P \) | Gas solubility |
Calculate molarity of a solution prepared by dissolving 18 g of NaCl (M = 58.5 g/mol) in 500 mL of solution.
Solution:
Moles of NaCl = \( \frac{18}{58.5} = 0.3077 \) mol
Volume of solution = 0.5 L
Molarity, \( M = \frac{0.3077}{0.5} = 0.6154 \, \text{mol/L} \)
What is the boiling point elevation of a solution made by dissolving 2 moles of a non-electrolyte in 1 kg of water (Kb for water = 0.512 °C/m)?
Solution:
Molality, \( m = \frac{2}{1} = 2 \, \text{mol/kg} \)
Boiling point elevation, \( \Delta T_b = K_b \times m = 0.512 \times 2 = 1.024^\circ C \)
New boiling point = 100 + 1.024 = 101.024 °C
Mastering solutions in chemistry will help you score well in JEE Mains and Advanced, as many questions revolve around concentration, colligative properties, and solubility equilibria.