Chapter 2: Solutions
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: Henry's law states that the solubility ($S$) of a gas in a liquid at a constant temperature is directly proportional to the partial pressure ($P$) of the gas present above the surface of the solution.
Mathematically: $S = K_H \cdot P$ (where $K_H$ is Henry's law constant).
Answer: Ebullioscopic constant (or Molal elevation constant) is defined as the elevation in boiling point produced when exactly 1 mole of a non-volatile, non-electrolyte solute is dissolved in 1 kg (1000 g) of the solvent.
Answer: Two or more solutions having the same osmotic pressure at a given constant temperature are called isotonic solutions.
Answer: The spontaneous and unidirectional flow of solvent molecules from a region of pure solvent to a solution (or from a dilute solution to a concentrated solution) through a semipermeable membrane is called osmosis.
2 Mark Questions (Short Answer-I)
Answer:
| Property | Ideal Solutions | Non-ideal Solutions |
|---|---|---|
| 1. Raoult's Law | Strictly obey Raoult's law over the entire range of concentrations. | Do not obey Raoult's law; show positive or negative deviations. |
| 2. Enthalpy Change | $\Delta H_{mix} = 0$ (No heat is absorbed or evolved). | $\Delta H_{mix} \neq 0$ (Heat is either absorbed or evolved). |
| 3. Volume Change | $\Delta V_{mix} = 0$ (Total volume is sum of individual volumes). | $\Delta V_{mix} \neq 0$ (Expansion or contraction occurs on mixing). |
| 4. Interactions | Solute-Solvent (A-B) interactions are identical to A-A and B-B interactions. | A-B interactions are stronger or weaker than A-A and B-B interactions. |
Answer:
- Evaporation is a surface phenomenon. In a pure solvent, the entire surface area is occupied by the volatile molecules of the solvent.
- When a non-volatile solute is added to form a solution, the solute particles occupy some of the surface area.
- This decreases the available surface area for the volatile solvent molecules to escape into the vapor phase.
- Consequently, the rate of evaporation decreases, which leads to a lowering of the vapor pressure compared to the pure solvent.
Answer:
Colligative properties are those physical properties of dilute solutions containing a non-volatile solute that depend only on the number of solute particles (molecules or ions) present in a given amount of solvent, and not on the chemical nature of the solute.
Any two examples:
- Relative lowering of vapor pressure.
- Elevation of boiling point.
- Depression of freezing point.
- Osmotic pressure.
3 Mark Questions (Short Answer-II / Numericals)
Answer:
1. Experimentally, for dilute solutions, the elevation of boiling point ($\Delta T_b$) is directly proportional to the molality ($m$) of the solution.
$$\Delta T_b \propto m \implies \Delta T_b = K_b \cdot m \quad \text{--- (Equation 1)}$$
Where $K_b$ is the ebullioscopic constant or molal elevation constant.
2. Molality ($m$) is defined as the number of moles of solute ($n_2$) dissolved per kilogram ($1000 \text{ g}$) of the solvent ($W_1$).
$$m = \frac{n_2}{W_1 (\text{in kg})} = \frac{W_2 / M_2}{W_1 / 1000}$$
$$m = \frac{1000 \cdot W_2}{M_2 \cdot W_1} \quad \text{--- (Equation 2)}$$
Where:
$W_2$ = mass of solute in grams.
$M_2$ = molar mass of solute in g/mol.
$W_1$ = mass of solvent in grams.
3. Substituting Equation 2 into Equation 1:
$$\Delta T_b = K_b \cdot \left[ \frac{1000 \cdot W_2}{M_2 \cdot W_1} \right]$$
4. Rearranging the equation to find the molar mass of the solute ($M_2$):
$$M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$$
This is the required relationship.
Answer:
Given Data:
Mass of solute ($W_2$) = $0.73 \text{ g}$
Mass of solvent/water ($W_1$) = $65 \text{ g}$
Boiling point of solution ($T_b$) = $100.15^\circ\text{C}$
Boiling point of pure solvent ($T_b^\circ$) = $100^\circ\text{C}$
Ebullioscopic constant ($K_b$) = $0.52 \text{ K kg mol}^{-1}$
Step 1: Calculate Elevation in Boiling Point ($\Delta T_b$)
$\Delta T_b = T_b - T_b^\circ = 100.15 - 100 = 0.15^\circ\text{C}$
(Note: The difference in temperature is the same in Celsius and Kelvin. Thus, $\Delta T_b = 0.15 \text{ K}$).
Step 2: Formula Application
We know, $$M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$$
$$M_2 = \frac{1000 \times 0.52 \times 0.73}{0.15 \times 65}$$
$$M_2 = \frac{379.6}{9.75}$$
$$M_2 = 38.93 \text{ g/mol}$$
The molar mass of the solute is $38.93 \text{ g mol}^{-1}$.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark numerical combined with a 1-mark theory question).
(b) Define: Cryoscopic constant ($K_f$). [1 Mark] March 2020
Answer (a): Numerical on Vapor Pressure Lowering
Given Data:
Vapor pressure of pure solvent ($P_1^\circ$) = $23.8 \text{ mm Hg}$
Mass of solute, urea ($W_2$) = $50 \text{ g}$
Molar mass of solute ($M_2$) = $60 \text{ g/mol}$
Mass of solvent, water ($W_1$) = $850 \text{ g}$
Molar mass of solvent ($M_1$) = $18 \text{ g/mol}$
Step 1: Calculate number of moles
Moles of urea ($n_2$) $= \frac{W_2}{M_2} = \frac{50}{60} = 0.833 \text{ mol}$
Moles of water ($n_1$) $= \frac{W_1}{M_1} = \frac{850}{18} = 47.22 \text{ mol}$
Step 2: Calculate mole fraction of solute ($x_2$)
$$x_2 = \frac{n_2}{n_1 + n_2} = \frac{0.833}{47.22 + 0.833} = \frac{0.833}{48.053} = 0.0173$$
By Raoult's law, the relative lowering of vapor pressure is exactly equal to the mole fraction of the solute.
$$\text{Relative lowering} = \frac{P_1^\circ - P_1}{P_1^\circ} = x_2 = 0.0173$$
Step 3: Calculate vapor pressure of the solution ($P_1$)
$$\frac{23.8 - P_1}{23.8} = 0.0173$$
$$23.8 - P_1 = 23.8 \times 0.0173 = 0.4117$$
$$P_1 = 23.8 - 0.4117 = 23.388 \text{ mm Hg}$$
Answers: Relative lowering is 0.0173. Vapor pressure of solution is 23.388 mm Hg.
Answer (b): Theory
Cryoscopic constant ($K_f$) or Molal depression constant is defined as the depression in freezing point produced when exactly 1 mole of a non-volatile, non-electrolyte solute is dissolved in 1 kg (1000 g) of the solvent. Its SI unit is $\text{K kg mol}^{-1}$.
(b) What is reverse osmosis? [1 Mark] Oct 2013, March 2019
Answer (a): Numerical on Isotonic Solutions
Given Data:
Solutions are Isotonic $\implies \pi_1 = \pi_2$
For cane sugar (Solution 1): 5% by mass means $W_1 = 5 \text{ g}$ in $100 \text{ mL}$ ($V_1$) of solution. Molar mass $M_1 = 342 \text{ g/mol}$.
For unknown solute (Solution 2): 0.877% by mass means $W_2 = 0.877 \text{ g}$ in $100 \text{ mL}$ ($V_2$) of solution. Molar mass $M_2 = ?$
Step 1: Apply Isotonic Condition Formula
Osmotic pressure $\pi = C R T = \frac{n}{V} R T = \frac{W}{M \cdot V} R T$
Since $\pi_1 = \pi_2$ at the same temperature, and $V_1 = V_2 = 100 \text{ mL}$:
$$\frac{W_1}{M_1 \cdot V_1} R T = \frac{W_2}{M_2 \cdot V_2} R T$$
$$\frac{W_1}{M_1} = \frac{W_2}{M_2}$$
Step 2: Calculation
$$\frac{5}{342} = \frac{0.877}{M_2}$$
$$M_2 = \frac{0.877 \times 342}{5} = \frac{299.934}{5}$$
$$M_2 = 59.98 \text{ g/mol} \approx 60 \text{ g/mol}$$
The molar mass of the unknown solute is $60 \text{ g/mol}$ (which is likely urea).
Answer (b): Theory
Reverse Osmosis: If a pressure larger than the osmotic pressure is applied to the solution side, the direction of osmosis is reversed. The pure solvent flows out of the solution through the semipermeable membrane into the pure solvent side. This process is called reverse osmosis (used in the desalination of seawater).
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