Solutions: Comprehensive NEET Chemistry Notes

1. Introduction to Solutions

Solutions are homogeneous mixtures composed of two or more substances. In a solution, the substance present in the greatest amount is typically called the solvent, while the substances present in lesser amounts are called solutes.

2. Types of Solutions

Solutions can be classified based on the physical state of the solute and solvent. Here are some common types:

Type of Solution	Solute	Solvent	Example
Gaseous	Gas	Gas	Air (oxygen in nitrogen)
Liquid	Gas	Liquid	Carbonated water (CO₂ in water)
Liquid	Liquid	Liquid	Alcohol in water
Solid	Liquid	Liquid	Salt in water
Solid	Solid	Solid	Alloys (e.g., brass)

Did You Know?

The human body relies on various solutions, such as blood plasma, to function properly.

3. Expressing Concentration of Solutions

The concentration of a solution can be expressed in several ways:

3.1 Mass Percentage (w/w)

The mass percentage of a component is given by: $Mass % of component = (\frac{Mass of component}{Total mass of solution}) \times 100$

3.2 Volume Percentage (v/v)

The volume percentage is: $Volume % of component = (\frac{Volume of component}{Total volume of solution}) \times 100$

3.3 Mass by Volume Percentage (w/v)

This is commonly used in medicine: $Mass/Volume % = (\frac{Mass of solute}{Volume of solution}) \times 100$

3.4 Parts Per Million (ppm)

Used for very dilute solutions: $ppm = (\frac{Number of parts of the component}{Total number of parts of all components}) \times 1 0^{6}$

3.5 Mole Fraction

The mole fraction of a component is: $x_{i} = \frac{n _{i}}{\sum n _{i}}$

3.6 Molarity (M)

Molarity is defined as: $M = \frac{Moles of solute}{Volume of solution in liters}$

3.7 Molality (m)

Molality is: $m = \frac{Moles of solute}{Mass of solvent in kg}$

NEET Tip:

Familiarize yourself with all concentration units, as questions in NEET can use any of these.

4. Solubility

Solubility is the maximum amount of solute that can dissolve in a specified amount of solvent at a specific temperature.

4.1 Solubility of Solids in Liquids

The solubility of solids typically increases with temperature.

4.2 Solubility of Gases in Liquids

The solubility of gases decreases with an increase in temperature but increases with pressure, described by Henry's law: $p = K_{H} x$

Real-life Application:

Scuba divers use Henry's law to avoid decompression sickness by controlling the pressure of gases in their breathing mix.

5. Vapor Pressure of Solutions

5.1 Raoult's Law

For a solution of volatile liquids, Raoult's law states:

$p_{1} = x_{1} p_{1}^{0}$

$p_{2} = x_{2} p_{2}^{0}$

$p_{total} = p_{1} + p_{2}$

5.2 Ideal and Non-Ideal Solutions

Ideal Solutions: Obey Raoult's law throughout the concentration range.
Non-Ideal Solutions: Show positive or negative deviations from Raoult's law.

6. Colligative Properties and Determination of Molar Mass

Colligative properties depend on the number of solute particles, not their identity.

6.1 Relative Lowering of Vapor Pressure

$\frac{Δ p _{1}}{p _{1}^{0}} = x_{2}$

6.2 Elevation of Boiling Point

$Δ T_{b} = K_{b} \cdot m$

6.3 Depression of Freezing Point

$Δ T_{f} = K_{f} \cdot m$

6.4 Osmosis and Osmotic Pressure

Osmotic pressure is given by: $Π = CRT$

NEET Problem-Solving Strategy:

Practice calculating molar masses using colligative properties.

Quick Recap

Solutions are homogeneous mixtures of solute and solvent.
Concentration can be expressed in various units.
Solubility depends on temperature and pressure.
Vapor pressure and colligative properties provide insights into solution behavior.

Practice Questions

Question: Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution. Solution: Moles of $Moles of NaOH = \frac{5}{40} = 0.125$ $Volume of solution in liters = 0.450$ $Molarity = \frac{0.125}{0.450} = 0.278 M$
Question: Determine the molality of a solution containing 2.5 g of ethanoic acid in 75 g of benzene. Solution: Moles of $Moles of C_{2} H_{4} O_{2} = \frac{2.5}{60} = 0.0417 Mass of benzene in kg = 0.075 Molality = \frac{0.0417}{0.075} = 0.556 mol/kg$
Question: Explain Henry’s law and its applications. Solution: Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Applications include carbonated beverages and scuba diving.
Question: Calculate the freezing point depression for a solution of 45 g of ethylene glycol in 600 g of water. Solution: $Moles of ethylene glycol = \frac{45}{62} = 0.726 Molality = \frac{0.726}{0.600} = 1.21 mol/kg Δ T_{f} = 1.86 \cdot 1.21 = 2.2 5^{\circ} C$
Question: Describe the process of osmosis. Solution: Osmosis is the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.