Chapter 2: Solutions
Complete Theory, Colligative Properties, Numericals & Fully Solved PYQs
1. Introduction and 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. The substance present in the larger proportion is the solvent, and the substance present in the smaller proportion is the solute.
Capacity of Solution to Dissolve Solute
- Saturated Solution: A solution in which no more solute can be dissolved at a given temperature. It represents a state of dynamic equilibrium between dissolved solute and undissolved solid solute.
- Unsaturated Solution: A solution containing less amount of solute than that required for saturation.
- Supersaturated Solution: A solution containing more solute than that in a saturated solution. These are highly unstable.
2. Solubility and Henry's Law
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. It is generally expressed in $\text{mol L}^{-1}$.
Factors Affecting Solubility
- Nature of solute and solvent: "Like dissolves like." Polar solutes dissolve in polar solvents (e.g., NaCl in water), and non-polar solutes dissolve in non-polar solvents (e.g., iodine in benzene).
- Effect of Temperature:
- If the dissolution process is endothermic ($\Delta H_{sol} > 0$), solubility increases with an increase in temperature (e.g., $KNO_3, KCl$).
- If the dissolution process is exothermic ($\Delta H_{sol} < 0$), solubility decreases with an increase in temperature (e.g., $Li_2SO_4$, gas in a liquid).
- Effect of Pressure (Henry's Law): Pressure has no significant effect on the solubility of solids and liquids. However, it greatly affects the solubility of gases in liquids.
Henry's Law
It states that the solubility ($S$) of a gas in a liquid is directly proportional to the pressure ($P$) of the gas over the solution at a given temperature.
$$ S \propto P \implies S = K_H \cdot P $$Where $K_H$ is Henry's law constant. Unit of $K_H$ is $\text{mol L}^{-1} \text{bar}^{-1}$.
Exception: Gases like $NH_3$ and $CO_2$ which react with water do not strictly obey Henry's law at all pressures.
3. Raoult's Law and Ideal Solutions
Raoult's Law for volatile liquids: It states that the partial vapor pressure of any volatile component of a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
For a binary solution of two volatile liquids 1 and 2:
$$ P_1 = P_1^0 \cdot x_1 \quad \text{and} \quad P_2 = P_2^0 \cdot x_2 $$Total vapor pressure $P = P_1 + P_2 = P_1^0 x_1 + P_2^0 x_2$.
Ideal and Non-Ideal Solutions
| Property | Ideal Solutions | Non-Ideal Solutions |
|---|---|---|
| Raoult's Law | Obeys strictly at all concentrations and temperatures. | Does not obey Raoult's law. |
| Enthalpy Change ($\Delta H_{mix}$) | Zero ($0$) | Not Zero ($\neq 0$) |
| Volume Change ($\Delta V_{mix}$) | Zero ($0$) | Not Zero ($\neq 0$) |
| Intermolecular Forces | A-B interactions are identical to A-A and B-B interactions. | A-B interactions are weaker or stronger than pure components. |
| Examples | Benzene + Toluene, n-Hexane + n-Heptane | Ethanol + Acetone, Chloroform + Acetone |
Non-Ideal Solutions show either:
- Positive Deviation: A-B forces are weaker than A-A/B-B. Vapor pressure is higher than expected. ($\Delta H_{mix} > 0, \Delta V_{mix} > 0$). Example: Ethanol + Acetone.
- Negative Deviation: A-B forces are stronger than A-A/B-B. Vapor pressure is lower than expected. ($\Delta H_{mix} < 0, \Delta V_{mix} < 0$). Example: Chloroform + Acetone.
4. Colligative Properties of Non-Electrolyte Solutions
Physical properties of solutions that depend only on the number of solute particles (molecules or ions) and not on their nature are called colligative properties. These apply to dilute solutions of non-volatile, non-electrolyte solutes.
I. Relative Lowering of Vapor Pressure
When a non-volatile solute is added to a solvent, the vapor pressure of the solvent decreases because solute particles block the surface area, reducing the rate of evaporation.
According to Raoult's Law for non-volatile solutes, the relative lowering of vapor pressure is equal to the mole fraction of the solute ($x_2$).
$$ \frac{P_1^0 - P_1}{P_1^0} = x_2 = \frac{n_2}{n_1 + n_2} $$For dilute solutions, $n_2 << n_1$, so $n_1 + n_2 \approx n_1$. Therefore: $$ \frac{\Delta P}{P_1^0} = \frac{W_2 \times M_1}{M_2 \times W_1} $$
($W_2, M_2$ = mass and molar mass of solute; $W_1, M_1$ = mass and molar mass of solvent).
II. Elevation of Boiling Point ($\Delta T_b$)
The boiling point of a solvent increases when a non-volatile solute is added because the vapor pressure is lowered. It requires a higher temperature to make the vapor pressure equal to atmospheric pressure.
$$ \Delta T_b = T_b - T_b^0 $$ $$ \Delta T_b = K_b \cdot m $$Where $K_b$ is the Ebullioscopic constant (Molal elevation constant) and $m$ is molality.
Molar Mass Relation: $$ M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1} $$
III. Depression of Freezing Point ($\Delta T_f$)
The freezing point of a solvent decreases upon the addition of a non-volatile solute.
$$ \Delta T_f = T_f^0 - T_f $$ $$ \Delta T_f = K_f \cdot m $$Where $K_f$ is the Cryoscopic constant (Molal depression constant).
Molar Mass Relation: $$ M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1} $$
IV. Osmotic Pressure ($\pi$)
Osmosis: The spontaneous, unidirectional flow of solvent molecules through a semipermeable membrane (SPM) from a pure solvent to a solution (or dilute to concentrated solution).
Osmotic Pressure: The excess hydrostatic pressure that must be applied to the solution side to prevent osmosis.
According to the van't Hoff equation for dilute solutions:
$$ \pi = C \cdot R \cdot T $$Where $C$ is molarity ($n_2/V$), $R$ is the gas constant ($0.08206 \text{ L atm K}^{-1}\text{mol}^{-1}$), and $T$ is temperature in Kelvin.
Molar Mass Relation: $$ M_2 = \frac{W_2 \cdot R \cdot T}{\pi \cdot V} $$
Types of Solutions based on Osmotic Pressure:
- Isotonic: Two solutions having the same osmotic pressure ($\pi_1 = \pi_2$).
- Hypertonic: A solution with a higher osmotic pressure than the other.
- Hypotonic: A solution with a lower osmotic pressure than the other.
5. Colligative Properties of Electrolytes (van't Hoff Factor)
Electrolytes undergo dissociation or association in solution, changing the number of particles. Colligative properties are thus observed to be abnormal (different from calculated values).
van't Hoff Factor ($i$)
$$ i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}} = \frac{\text{Theoretical Molar Mass}}{\text{Observed Molar Mass}} $$
Modified equations for electrolytes:
- $\frac{\Delta P}{P_1^0} = i \cdot x_2$
- $\Delta T_b = i \cdot K_b \cdot m$
- $\Delta T_f = i \cdot K_f \cdot m$
- $\pi = i \cdot C \cdot R \cdot T$
Relation between $i$ and Degree of Dissociation ($\alpha$)
Degree of dissociation ($\alpha$) is the fraction of total moles of the electrolyte that dissociate.
$$ \alpha = \frac{i - 1}{n - 1} \implies i = 1 + \alpha(n - 1) $$Where $n$ is the number of moles of ions produced from one mole of the electrolyte (e.g., for $NaCl, n=2$; for $BaCl_2, n=3$).
6. Solved Textbook Numericals
Problem 1: Henry's Law
The solubility of $N_2$ gas in water at 25°C and 1 bar is $6.85 \times 10^{-4} \text{ mol L}^{-1}$. Calculate Henry's law constant and the molarity of $N_2$ gas dissolved in water under atmospheric conditions when partial pressure of $N_2$ in atmosphere is 0.75 bar.
Solution:
Step 1: Calculate $K_H$
$S = K_H \cdot P \implies K_H = \frac{S}{P}$
$K_H = \frac{6.85 \times 10^{-4} \text{ mol L}^{-1}}{1 \text{ bar}} = 6.85 \times 10^{-4} \text{ mol L}^{-1} \text{bar}^{-1}$
Step 2: Calculate Solubility at 0.75 bar
$S = K_H \cdot P$
$S = (6.85 \times 10^{-4}) \times (0.75) = 5.138 \times 10^{-4} \text{ mol L}^{-1}$
Answer: $K_H = 6.85 \times 10^{-4} \text{ mol L}^{-1}\text{bar}^{-1}$; Molarity = $5.138 \times 10^{-4} \text{ M}$.
Problem 2: Boiling Point Elevation
The normal boiling point of ethyl acetate is 77.06 °C. A solution of 50 g of a non-volatile solute in 150 g of ethyl acetate boils at 84.27 °C. Evaluate the molar mass of solute if $K_b$ for ethyl acetate is $2.77 \text{ K kg mol}^{-1}$.
Solution:
Given: $W_2 = 50 \text{ g}, W_1 = 150 \text{ g}, K_b = 2.77 \text{ K kg mol}^{-1}$
$T_b^0 = 77.06^\circ\text{C}, \quad T_b = 84.27^\circ\text{C}$
$\Delta T_b = T_b - T_b^0 = 84.27 - 77.06 = 7.21^\circ\text{C} = 7.21 \text{ K}$ (Difference remains same in K)
Using formula: $M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$
$$ M_2 = \frac{1000 \times 2.77 \times 50}{7.21 \times 150} = \frac{138500}{1081.5} \approx 128 \text{ g mol}^{-1} $$
Answer: The molar mass of the solute is $128 \text{ g mol}^{-1}$.
7. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. State Henry's Law. (March 2013, Oct 2015, March 2022)
Answer: Henry's law states that the solubility ($S$) of a gas in a liquid at constant temperature is directly proportional to the pressure ($P$) of the gas above the solution. mathematically, $S = K_H \cdot P$.
Q2. Define: Ebullioscopic constant. (March 2016, July 2018)
Answer: Ebullioscopic constant (Molal elevation constant, $K_b$) is defined as the elevation in boiling point produced when 1 mole of a non-volatile solute is dissolved in 1 kg (1000 g) of the solvent.
Q3. What are isotonic solutions? (March 2019, Oct 2021)
Answer: Two or more solutions having the same osmotic pressure at a given temperature are called isotonic solutions.
2 Mark Questions (SA-I)
Q4. Distinguish between ideal and non-ideal solutions. (March 2015, Oct 2017)
- Ideal Solutions: Obey Raoult's law over entire range of concentration. $\Delta H_{mix} = 0$, $\Delta V_{mix} = 0$. Solute-solvent interactions are similar to solute-solute and solvent-solvent interactions.
- Non-ideal Solutions: Do not obey Raoult's law. $\Delta H_{mix} \neq 0$, $\Delta V_{mix} \neq 0$. Solute-solvent interactions are different from pure component interactions.
Q5. Why is the vapor pressure of a solution containing a non-volatile solute lower than that of the pure solvent? (March 2014, March 2020)
Answer: Evaporation is a surface phenomenon. In a pure solvent, the entire surface is occupied by volatile solvent molecules. When a non-volatile solute is added, a part of the surface is occupied by non-volatile solute particles. This decreases the available surface area for solvent molecules to escape, thus lowering the rate of evaporation and consequently lowering the vapor pressure.
3 Mark Questions (SA-II)
Q6. Derive the relationship between elevation in boiling point and molar mass of a non-volatile solute. (Oct 2014, March 2017, March 2023)
1. We know that elevation in boiling point ($\Delta T_b$) is directly proportional to molality ($m$) of the solution:
$\Delta T_b \propto m \implies \Delta T_b = K_b \cdot m$ --- (Equation 1)
Where $K_b$ is the boiling point elevation constant.
2. The molality ($m$) is the number of moles of solute ($n_2$) per kg of solvent ($W_1$ in kg).
$m = \frac{n_2}{W_1(\text{in kg})} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \cdot W_2}{M_2 \cdot W_1}$ --- (Equation 2)
Where $W_2$ = mass of solute (g), $M_2$ = molar mass of solute (g/mol), $W_1$ = mass of solvent (g).
3. Substituting (Equation 2) into (Equation 1):
$\Delta T_b = K_b \cdot \frac{1000 \cdot W_2}{M_2 \cdot W_1}$
4. Rearranging for $M_2$:
$$ M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1} $$
This is the required relationship.
Q7. A solution containing 0.73 g of a solute in 65 g of water boils at 100.15 °C. Calculate the molar mass of the solute. ($K_b$ for water = 0.52 K kg/mol, Boiling point of pure water = 100 °C) (March 2018, Oct 2020)
Given:
$W_2 = 0.73 \text{ g}$ (mass of solute)
$W_1 = 65 \text{ g}$ (mass of solvent)
$T_b = 100.15^\circ\text{C}$, $T_b^0 = 100^\circ\text{C}$
$K_b = 0.52 \text{ K kg mol}^{-1}$
Step 1: Calculate $\Delta T_b$
$\Delta T_b = T_b - T_b^0 = 100.15 - 100 = 0.15^\circ\text{C} = 0.15 \text{ K}$
Step 2: Use formula
$M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$
$M_2 = \frac{1000 \times 0.52 \times 0.73}{0.15 \times 65}$
$M_2 = \frac{379.6}{9.75} = 38.93 \text{ g mol}^{-1}$
Final Answer: The molar mass of the solute is $38.93 \text{ g/mol}$.
📚 Also Read
Lecture Notes🏆 Complete Maharashtra HSC Class 12 Chemistry Preparation
Prepare for the Maharashtra HSC Class 12 Chemistry Board Exam with chapter-wise revision notes, important questions, PYQs, formula sheets, mock tests, quick revision resources and exam-oriented study material. Everything you need to score high in one comprehensive learning hub.
🚀 Explore the Complete Maharashtra HSC Chemistry Hub
No comments:
Post a Comment