Unit 1: Solutions

1.1 Types of Solutions

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

• Solute: Component present in smaller quantity.
• Solvent: Component present in larger quantity.

Solutions can be classified based on the physical state of the solvent. Our primary focus is on liquid solutions.

1.2 Expression of Concentration of Solutions

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

• Mass Percentage (w/w): \( \text{Mass } % = \frac{\text{Mass of component in solution}}{\text{Total mass of solution}} \times 100 \)
• Volume Percentage (v/v): \( \text{Volume } % = \frac{\text{Volume of component}}{\text{Total volume of solution}} \times 100 \)
• Molarity (M): \( M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} \)
• Molality (m): \( m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \)
• Mole Fraction (x): \( x_A = \frac{n_A}{n_A + n_B} \)

1.3 Solubility of Gases in Liquids

The solubility of gases in liquids is greatly affected by pressure and temperature.

Henry’s Law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution. \[ p = K_H \cdot x \] where \( p \) is the partial pressure of the gas, \( x \) is the mole fraction of the gas in solution, and \( K_H \) is Henry’s law constant.

1.4 Raoult’s Law and Solid Solutions

For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. \[ p_1 = p_1^0 x_1 \] where \( p_1^0 \) is the vapour pressure of pure component 1 at the same temperature.

An Ideal Solution strictly obeys Raoult’s law over the entire range of concentration. Most solutions, however, are non-ideal and show either positive or negative deviation from Raoult’s law.

1.5 Colligative Properties

Properties of dilute solutions which depend only on the number of solute particles (molecules/ions) and not on their nature.

1. Relative Lowering of Vapour Pressure

When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lower than that of the pure solvent. \[ \frac{p_1^0 - p_1}{p_1^0} = x_2 \]

2. Elevation of Boiling Point (\( \Delta T_b \))

The boiling point of a solution containing a non-volatile solute is higher than that of the pure solvent. \[ \Delta T_b = K_b \cdot m \] where \( K_b \) is the molal elevation constant (ebullioscopic constant) and \( m \) is molality.

3. Depression of Freezing Point (\( \Delta T_f \))

The freezing point of a solution is lower than that of the pure solvent. \[ \Delta T_f = K_f \cdot m \] where \( K_f \) is the molal depression constant (cryoscopic constant).

4. Osmosis and Osmotic Pressure (\( \pi \))

Osmosis is the spontaneous flow of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semi-permeable membrane. The osmotic pressure is the excess pressure that must be applied to a solution to prevent osmosis. \[ \pi = C R T \] where \( C \) is the molar concentration (molarity) of the solution, \( R \) is the gas constant, and \( T \) is temperature.

1.6 Abnormal Molecular Masses and van’t Hoff Factor

When a solute undergoes dissociation or association in a solution, the observed molecular mass differs from the normal (expected) molecular mass. The van’t Hoff factor (\( i \)) accounts for the extent of dissociation or association.

\[ i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} \] \[ i = \frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}} \]

Modified Colligative Properties Equations:

• Relative lowering of vapour pressure: \( \frac{p_1^0 - p_1}{p_1^0} = i \cdot x_2 \)
• Elevation of Boiling Point: \( \Delta T_b = i \cdot K_b \cdot m \)
• Depression of Freezing Point: \( \Delta T_f = i \cdot K_f \cdot m \)
• Osmotic Pressure: \( \pi = i \cdot C R T \)

Competency-Based Questions (CBQs)

Q1. (CBSE 2023) Based on intermolecular forces and Raoult’s law, a solution of chloroform (\(CHCl_3\)) and acetone (\(CH_3COCH_3\)) shows a specific deviation. Identify the type of deviation and provide the underlying reason.

Answer: The solution of chloroform and acetone shows a negative deviation from Raoult’s law.

Reason: In pure state, both chloroform and acetone have dipole-dipole interactions. When they are mixed, the hydrogen atom of chloroform forms a hydrogen bond with the highly electronegative oxygen atom of acetone: \[ CH_3COCH_3 \cdots H-CCl_3 \] Due to this new intermolecular hydrogen bonding, the interactions between chloroform and acetone molecules are stronger than the pure A-A and B-B interactions. This decreases the escaping tendency of molecules, leading to a lower vapour pressure than predicted by Raoult’s law, and hence a negative deviation.

Q2. (Sample Paper 2023) A 5% (by mass) solution of cane sugar (molar mass 342 g/mol) in water has a freezing point of 271 K. Calculate the freezing point of a 5% (by mass) solution of glucose (molar mass 180 g/mol) in water if the freezing point of pure water is 273.15 K.

Answer: It is given that mass percentage of both solutions is 5%. This means 5 g of solute is present in 100 g of solution, i.e. mass of solvent (water) = 95 g.

For cane sugar solution:

• \( w_2 \) (cane sugar) = 5 g
• \( M_2 \) = 342 g/mol
• \( w_1 \) (water) = 95 g
• \( \Delta T_f \) = Freezing point of pure water - Freezing point of solution = 273.15 - 271 = 2.15 K

Using the formula: \[ \Delta T_f = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1} \] \[ 2.15 = K_f \cdot \frac{5 \times 1000}{342 \times 95} \] \[ K_f = \frac{2.15 \times 342 \times 95}{5000} = 13.99 , \text{K kg mol}^{-1} \]

For glucose solution:

• \( w_2 \) (glucose) = 5 g
• \( M_2 \) = 180 g/mol
• \( w_1 \) (water) = 95 g
• \( K_f \) = 13.99 K kg/mol

\[ \Delta T_f’ = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1} \] \[ \Delta T_f’ = 13.99 \cdot \frac{5 \times 1000}{180 \times 95} = 4.09 , \text{K} \]

The freezing point of the glucose solution = Freezing point of pure water - \(\Delta T_f’\) \[ T_f’ = 273.15 - 4.09 = 269.06 , \text{K} \]

Q3. (CBSE 2020) Why is osmotic pressure considered the preferred colligative property for determining the molecular masses of macromolecules such as proteins and polymers?

Answer: Osmotic pressure is preferred for determining the molecular masses of macromolecules for the following reasons:

1. Magnitude: The osmotic pressure values are reasonably large and measurable even for very dilute solutions of macromolecules. In contrast, the changes in boiling point or freezing point are negligibly small.
2. Temperature: Osmotic pressure is measured at room temperature, which prevents the denaturation or degradation of sensitive biomolecules like proteins that might decompose at higher temperatures (e.g., during boiling point elevation).
3. Molarity vs Molality: Osmotic pressure depends on molarity instead of molality. For large polymers, it is easier to prepare solutions of known volume than known mass of solvent accurately.