Class 12 Chemistry Chapter 1 Solution

Key Concept: Raoult’s Law Application

b) 60 mmHg
[Solution Description] According to Raoult’s law, the vapour pressure of the solution $(P_{\text{solution}})$ is proportional to the mole fraction of the solvent $(\chi_{\text{solvent}})$. The formula is $$P_{\text{solution}} = P^0_{\text{solvent}} \times \chi_{\text{solvent}}$$, where $$P^0_{\text{solvent}}$$ is the vapour pressure of the pure solvent.

Given:
Number of moles of solute = 1
Number of moles of solvent = 3
$P^0_{\text{solvent}}$ = 80 mmHg

First, calculate $\chi_{\text{solvent}}$:
$$
\chi_{\text{solvent}} = \frac{\text{moles of solvent}}{\text{total moles}} = \frac{3}{3 + 1} = \frac{3}{4} = 0.75
$$

Now calculate $$P_{\text{solution}}$$:
$$
P_{\text{solution}} = 80 \times 0.75 = 60 \text{ mmHg}
$$
Therefore, the vapour pressure of the solution is 60 mmHg.

Click Here to Download Notes

Key Concept: Multi-Step Problem Solving, Real-World Scenario Analysis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To analyze the assertion and reason:

– Assertion: Colligative properties such as elevation of boiling point or depression of freezing point are affected by the number of particles present in a solution. The van’t Hoff factor (i) accounts for the degree of association or dissociation and is necessary for accurately calculating these properties. Therefore, it indeed plays a crucial role when solutes like ethanoic acid partially associate in solvents like benzene.

– Reason: When ethanoic acid is dissolved in benzene, it tends to form dimers due to hydrogen bonding. This dimerization causes the observed particle count to decrease, affecting $i$. For instance, if $i \approx 0.5$, it indicates significant dimerization (as fewer moles of particles are present than expected if no association occurred). Hence, $i$ helps quantify this association.

Both Assertion and Reason are true statements. Furthermore, the Reason correctly explains why the Assertion holds true, especially in scenarios involving association like dimerization.

Click Here to Download Notes

Key Concept: Application of Henry’s Law

c) 800 Pa
[Solution Description] According to Henry’s Law, the partial pressure $p$ of a gas above the solution is given by the formula $$ p = KH \cdot x $$. Here, $ KH = 8.0 \times 10^4 \, \text{Pa} $ and $ x = 0.01 $.

Calculate $ p $:
$$
p = 8.0 \times 10^4 \times 0.01 = 8.0 \times 10^2 \, \text{Pa}
$$

Thus, the partial pressure of nitrogen gas is $800 \, \text{Pa}$.

Click Here to Download Notes

Key Concept: Dynamic Equilibrium

d) The rate of dissolution equals the rate of crystallization.
[Solution Description] In a saturated solution at dynamic equilibrium, the rate of dissolution of the solute equals the rate of crystallization. This means that while individual molecules might be moving between phases, the overall concentration of the solute remains constant.

Click Here to Download Notes

Key Concept: Osmotic Pressure and Colligative Properties, Advanced Calculation, Reverse Osmosis and Environmental Applications

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion that “osmotic pressure is inversely proportional to the volume of solute” is incorrect. According to the formula for osmotic pressure $P = CRT$, osmotic pressure directly depends on the concentration (C) of the solution, not the volume of solute itself. Concentration is defined as moles of solute per volume of solution, thus indirectly considering volume but not in an inverse manner.

On the other hand, the reason statement regarding reverse osmosis is accurate. In reverse osmosis, pressure applied must exceed the natural osmotic pressure of the solution to drive solvent molecules from solution back to pure solvent, effectively removing impurities.

Since the assertion is false and reason is true, the correct option is d).

Click Here to Download Notes

Key Concept: Deviations and Molecular Forces, Complex Interactions

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion is true as an ideal solution follows Raoult’s Law with an enthalpy change of zero due to equal strength of all types of molecular interactions, i.e., solute-solute, solvent-solvent, and solute-solvent. However, the reason is false. In reality, positive deviation from Raoult’s law happens when the interactions between unlike molecules (solute-solvent) are weaker than the interactions between the similar molecules (solute-solute or solvent-solvent). This results in higher vapor pressure than predicted by Raoult’s law, not stronger interactions as mentioned in the Reason.

Click Here to Download Notes

Key Concept: Industrial Application, Dynamic Equilibrium Analysis

b) $CO_2$ escapes until new equilibrium is reached.
[Solution Description] In the sealed system, dynamic equilibrium exists: the rate of $CO_2$ entering the solution equals the rate leaving it. Upon opening, the pressure drops to 1 atm, decreasing the gas solubility by:

\[
p = KH \cdot x \implies x = \frac{1}{3.3 \times 10^3} = 3.03 \times 10^{-4}
\]

At lower pressure, less $CO_2$ dissolves. The excess $CO_2$ escapes, maintaining equilibrium with fewer molecules in solution.

Click Here to Download Notes

Key Concept: Complex Deviations, Advanced Colligative Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] Both assertion and reason are true but they address different aspects of solutions. Negative deviation occurs when the interactions between solute and solvent molecules are stronger than those in the pure components. This causes the vapor pressure to be lower than expected by Raoult’s law. On the other hand, colligative properties depend on the addition of solute particles which affects properties like boiling point elevation due to change in vapor pressure. While both statements are accurate, they do not explain each other directly.

Click Here to Download Notes

Key Concept: Simple Calculation

b) 2 ppm
[Solution Description] To find the ppm, use the formula:

$$
\text{ppm} = \frac{\text{mass of solute} \times 10^6}{\text{total mass of solution}}
$$

Substituting the given values:

$$
\text{ppm} = \frac{0.002 \times 10^6}{1000} = 2 \, \text{ppm}
$$

Thus, the concentration is 2 ppm.

Click Here to Download Notes

Key Concept: Comparative Analysis

b) Toluene
[Solution Description] To find the mole fraction, we use the formula $$x_i = \frac{\text{moles of } i}{\text{total moles}}$$. So, for benzene, $$x_{\text{benzene}} = \frac{1.5}{1.5 + 3.5} = \frac{1.5}{5} = 0.3$$. For toluene, $$x_{\text{toluene}} = \frac{3.5}{1.5 + 3.5} = \frac{3.5}{5} = 0.7$$. Therefore, toluene has the higher mole fraction.

Click Here to Download Notes

Key Concept: Unit Identification

c) Moles per kilogram

[Solution Description]

Molality is defined as the number of moles of solute per kilogram of solvent. Thus, its unit is moles per kilogram (mol/kg).

Click Here to Download Notes

Key Concept: Reverse Osmosis Process

a) Water flows into the solution

[Solution Description]

In reverse osmosis, if the applied pressure is less than the osmotic pressure, the natural osmotic flow will dominate. This means the solvent (water) will move from the side with lower solute concentration to the side with higher solute concentration through the semipermeable membrane, rather than being forced in the opposite direction.

Click Here to Download Notes

Key Concept: Vapour Pressure

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion is true because the mixture of ethanol and acetone exhibits weaker interactions between the different molecules (A-B interactions) compared to the interactions within each pure component (A-A and B-B interactions). As a result, it leads to increased vapour pressure compared to what Raoult’s law predicts, demonstrating positive deviation. The reason correctly explains why the assertion is true.

Click Here to Download Notes

Key Concept: Osmosis Application

d) The strip shrinks due to loss of water.
[Solution Description] When a potato strip is immersed in a concentrated sugar solution, water moves out from the potato cells to the sugar solution through osmosis. This causes the potato strip to lose water and become limp or shrink. Osmosis is the process where water moves across a semi-permeable membrane from a region of lower solute concentration to higher solute concentration.

Click Here to Download Notes

Key Concept: Simple Calculation

b) 0.25 M
[Solution Description]
To find molarity, we use the formula:

$$
M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}
$$

Here, moles of solute = 1 and volume of solution = 4 litres.

So,

$$
M = \frac{1}{4} = 0.25 \, \text{mol/L}
$$

Click Here to Download Notes

Key Concept: Association and Dissociation Effects, Osmotic Pressure in Biomolecules

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that osmotic pressure is a suitable method for determining molar mass of biomolecules. This is true as many biomolecules are sensitive to heat, which makes methods requiring elevated temperatures unsuitable. The reason highlights that association in solutions affects osmotic pressure, leading to lower observed values than expected if dissociation was present. While both statements are true, they address different aspects of colligative properties: one about measurement conditions (Assertion) and the other about effects on colligative measures (Reason). Thus, Reason is not the correct explanation of Assertion.

Click Here to Download Notes

Key Concept: Multi-step Calculation

b) 10%
[Solution Description] To calculate the mass percentage, we use the formula:
$$
\text{Mass %} = \left( \frac{\text{Mass of component}}{\text{Total mass of solution}} \right) \times 100
$$
Here, the mass of the component (salt) is 20 g and the total mass of the solution is the sum of the masses of salt and water:
$$
\text{Total mass of solution} = 20 \, \text{g} + 180 \, \text{g} = 200 \, \text{g}
$$
Plugging these values into the formula gives:
$$
\text{Mass %} = \left( \frac{20 \, \text{g}}{200 \, \text{g}} \right) \times 100 = 10\%
$$

Click Here to Download Notes

Key Concept: Calculation with Conversion

c) 3.472 mol/kg
[Solution Description]
First, calculate the moles of $CH_3OH$ using its molar mass:
$$
\text{Moles of CH}_3\text{OH} = \frac{20 \text{ g}}{32 \text{ g/mol}} = 0.625 \text{ mol}
$$
Next, convert the mass of water to kilograms:
$$
180 \text{ g} = 0.180 \text{ kg}
$$
Use the molality formula:
$$
m = \frac{0.625 \text{ mol}}{0.180 \text{ kg}} \approx 3.472 \text{ mol/kg}
$$
Thus, the molality is approximately 3.472 mol/kg.

Click Here to Download Notes

Key Concept: Van’t Hoff Factor

c) Greater than 1

[Solution Description]

The van’t Hoff factor (i) indicates the degree of dissociation of a solute in a solution. For complete dissociation, the van’t Hoff factor will be greater than one since more ions or particles are produced per formula unit of solute dissolved.

Click Here to Download Notes

Key Concept: Concept Application

c) 2.15 K
[Solution Description]
The change in boiling point $\Delta T_b$ is calculated by subtracting the boiling point of the pure solvent from that of the solution:
$
\Delta T_b = 375.15 \text{ K} – 373 \text{ K} = 2.15 \text{ K}
$

Therefore, the change in boiling point is 2.15 K.

Click Here to Download Notes

Key Concept: Complex Problem Solving, Multi-Step Calculation

a) 4.00

[Solution Description]

To find the van’t Hoff factor $i$, we use the formula for freezing point depression:

$i = \frac{\Delta T_f}{K_f m}$

$$i = \frac{3.72}{1.86 \times 0.5} = \frac{3.72}{0.93} \approx 4.00$$

Therefore, the van’t Hoff factor $i$ is approximately $4.00$.

Click Here to Download Notes

Key Concept: Mole Fraction Calculation

b) 0.134
[Solution Description]
The number of moles of acetone, $n_{\text{acetone}}$, is calculated as:
$$
n_{\text{acetone}} = \frac{50 \, \text{g}}{58 \, \text{g/mol}} = 0.8621 \, \text{mol}
$$

The number of moles of water, $n_{\text{water}}$, is calculated as:
$$
n_{\text{water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.5556 \, \text{mol}
$$

The total number of moles, $n_{\text{total}}$, is:
$$
n_{\text{total}} = n_{\text{acetone}} + n_{\text{water}} = 0.8621 + 5.5556 = 6.4177 \, \text{mol}
$$

The mole fraction of acetone, $x_{\text{acetone}}$, is given by:
$$
x_{\text{acetone}} = \frac{n_{\text{acetone}}}{n_{\text{total}}} = \frac{0.8621}{6.4177} = 0.1343
$$

Click Here to Download Notes

Key Concept: Mass Percentage Calculation

b) 20%
[Solution Description]
To find the mass percentage, we use the formula:
$$
\text{Mass percentage} = \left(\frac{\text{mass of solute}}{\text{mass of solute} + \text{mass of solvent}}\right) \times 100
$$
Here, the mass of solute (sodium chloride) is 25 g, and the mass of solvent (water) is 100 g. Thus,
$$
\text{Mass percentage} = \left(\frac{25}{25 + 100}\right) \times 100 = \left(\frac{25}{125}\right) \times 100 = 20%
$$
Hence, the mass percentage of sodium chloride is 20%.

Click Here to Download Notes

Key Concept: Raoult’s Law Application

b) $\Delta p_1 = x_2 p_1^0$
[Solution Description] Raoult’s Law states that the reduction in vapour pressure $\Delta p_1$ can be given by $\Delta p_1 = x_2 p_1^0$, where $x_2$ is the mole fraction of the solute and $p_1^0$ is the vapour pressure of the pure solvent.

Click Here to Download Notes

Key Concept: Concentration Comparison

d) Volume percentage
[Solution Description]
Wine generally contains a small percentage of alcohol mixed with a relatively large amount of water. For solutions where one component is present in much smaller amounts than others, volume percentage is often used to express concentration because it directly relates to the actual liquid volumes involved:

Volume percentage is calculated as:

$$
\%\text{Volume} = \left(\frac{\text{volume of solute}}{\text{total volume of solution}}\right) \times 100
$$

It gives a clear indication of how much of the substance contributes to the overall composition of the liquid mixture.

Click Here to Download Notes

Key Concept: Theoretical Analysis, Multi-Concept Integration

b) Greater for sodium chloride solution due to ionization
[Solution Description]
In this problem, we analyze how different solutes influence the lowering of vapour pressure based on their nature.

Glucose is a non-electrolyte and does not dissociate into ions in the solution, whereas sodium chloride $(\text{NaCl})$ is an electrolyte that dissociates completely into two ions $(\text{Na}^+ and \text{Cl}^-)$.

The colligative properties, including lowering of vapour pressure, depend on the number of solute particles. For solutions of equal concentration, the effective number of particles differs:

– $\text{NaCl}$ produces two particles per formula unit, doubling the effect on vapour pressure compared to glucose which remains as one particle per molecule because it does not dissociate.

Consequently, the lowering of vapour pressure will be greater in the sodium chloride solution due to the ionization factor impacting the number of solute particles present.

Click Here to Download Notes

Key Concept: Ideal Solution

b) Obeys Raoult’s Law at all concentrations
[Solution Description]
An ideal solution adheres to Raoult’s Law at all concentrations, exhibiting no changes in enthalpy or volume upon mixing, i.e., $$Δ_{\text{mix}}H = 0$$ and $$Δ_{\text{mix}}V = 0$$. This means that it behaves predictably according to Raoult’s Law without any deviations.

Click Here to Download Notes

Key Concept: Raoult’s Law Application

b) 80 mmHg
[Solution Description] According to Raoult’s Law, the vapour pressure of the solution $p_1$ can be calculated using the formula:
$
p_1 = x_1 \times p_1^0
$
Substituting the given values:
$
p_1 = 0.8 \times 100 = 80 \text{ mmHg}
$

Click Here to Download Notes

Key Concept: Common Usage

c) Medicine
[Solution Description] Mass by volume percentage is a unit that is commonly used in medicine and pharmacy to describe the concentration of solutions, particularly those administered intravenously or in liquid medications.

Click Here to Download Notes

Key Concept: Raoult’s Law

b) Mole fraction of the solvent
[Solution Description] Raoult’s Law states that the vapour pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent, given by $ p_1 = x_1 p^0_1 $, where $ x_1 $ is the mole fraction of the solvent.

Click Here to Download Notes

Key Concept: Basic Concept

b) Vapour pressure decreases.
[Solution Description] Adding a non-volatile solute to a solvent reduces its vapour pressure because the solute particles occupy space at the surface, thereby reducing the number of solvent molecules available to escape into the vapour phase.

Click Here to Download Notes

Key Concept: Calculation with Conversion

b) 20%
[Solution Description] The formula for volume percentage is:
$
\text{Volume $%$} = \left( \frac{\text{Volume of component}}{\text{Total volume of solution}} \right) \times 100
$
Substituting the given values:
$\text{Volume $%$} = \left( \frac{60 \, \text{mL}}{300 \, \text{mL}} \right) \times 100 = 20%$
Therefore, the volume percentage is 20%.

Click Here to Download Notes

Key Concept: Ideal Solution Rarity, Thermodynamic Implications

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that ideal solutions serve as a useful standard in thermodynamics for simplifying calculations related to enthalpy and volume. This is true because when dealing with chemical reactions and mixtures, having quantities like the enthalpy of mixing and the volume of mixing equal to zero eliminates additional variables that could complicate calculations.

The reason given supports this by explaining why those calculations are simplified: the absence of enthalpy and volume changes during mixing in ideal solutions simplifies theoretical models and predictions in thermodynamics. Hence, both the Assertion and Reason are true, and the Reason accurately explains the Assertion.

Click Here to Download Notes

Key Concept: Definition Check

c) The vapor pressure is higher than expected according to Raoult’s Law.
[Solution Description] In a solution showing positive deviation, the A-B interactions are weaker than A-A and B-B interactions. This causes an increase in the vapor pressure compared to what is predicted by Raoult’s Law.

Click Here to Download Notes

Key Concept: Dynamic Equilibrium

d) Rate of dissolution equals rate of crystallization
[Solution Description] Dynamic equilibrium is achieved when the rate of dissolution equals the rate of crystallization. At this point, the concentration of dissolved solute remains constant.

Click Here to Download Notes

Key Concept: Basic Concept

b) 3
[Solution Description]
The van’t Hoff factor (i) represents the number of particles a solute forms in solution. If a solute completely dissociates, each molecule breaks into multiple particles. For example, if 1 mole of a solute dissociates into 3 ions, then $i = 3$. Therefore, for complete dissociation into three ions, the van’t Hoff factor is 3.

Click Here to Download Notes

Key Concept: Real-World Problem

c) Due to decreased solubility of oxygen in blood.
[Solution Description] At high altitudes, the atmospheric pressure is lower than at sea level. According to Henry’s Law, solubility decreases with a decrease in partial pressure. Therefore, less oxygen dissolves in the blood at higher altitudes, making it harder for climbers to get the necessary oxygen.

Click Here to Download Notes

Key Concept: Solvent-Solute Match

a) Sodium chloride $(NaCl)$
[Solution Description] Water is a polar solvent, which means it can best dissolve polar solutes due to the “like dissolves like” principle. Among the options provided, sodium chloride (NaCl) is an ionic compound which behaves as a polar solute. Hence, it will dissolve in water.

Click Here to Download Notes

Key Concept: Temperature and KH

a) It increases
[Solution Description] As temperature increases, the solubility of gases typically decreases because dissolution is exothermic. This means that the Henry’s Law constant $KH$, which represents the inverse relationship between solubility and pressure, usually increases with temperature since solubility decreases.

Click Here to Download Notes

Key Concept: Colligative Properties

a) Boiling point elevation
[Solution Description]
Colligative properties depend on the number of solute particles but not their identity. When a non-volatile solute is added to a solvent, the boiling point elevation occurs because the addition of solute lowers the vapor pressure, thus requiring a higher temperature to reach the atmospheric pressure.

This property is known as boiling point elevation.

Click Here to Download Notes

Key Concept: Saturated Solution

b) Saturated

[Solution Description] A saturated solution contains the maximum amount of solute that can be dissolved at a specific temperature and pressure, meaning no more solute can be dissolved.

Click Here to Download Notes

Key Concept: Binary Solutions

b) Two

[Solution Description] A binary solution is composed of exactly two components: one solute and one solvent.

Click Here to Download Notes

Key Concept: Relative Lowering

a) 0.063

[Solution Description]

Using the formula for relative lowering of vapour pressure:

$\frac{p_1^0 – p}{p_1^0} = x_2$

where:

– $p_1^0 = 23.8$ mm Hg (pure water)

– $p = 22.3$ mm Hg (solution)

Substitute these values to find $x_2$ (mole fraction of solute):

$\frac{23.8 – 22.3}{23.8} = x_2$

Simplifying,

$\frac{1.5}{23.8} = x_2$

Calculating,

$x_2 \approx 0.063$

Therefore, the mole fraction of the solute is approximately 0.063.

Click Here to Download Notes

Key Concept: Azeotropic Mixtures, Non-Ideal Solutions

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

In azeotropic mixtures, the composition of vapor is identical to that of the liquid phase at a specific temperature and pressure, resulting in a constant boiling point. This behavior leads to no further separation through regular distillation processes. Therefore, Assertion is true as azeotropes cannot be fully separated by simple distillation.

The Reason states that non-ideal solutions exhibit deviations from Raoult’s Law because of varying interactions between solute and solvent molecules. These interactions can either be stronger or weaker compared to pure components interacting with themselves, causing positive or negative deviations.

Although both statements are accurate individually, the reason does not directly explain why azeotropic mixtures behave as described in the assertion. Azeotropes form due to unique interactive properties leading to an equilibrium state where further separation isn’t feasible rather than solely because of general non-ideality.

Thus, while both Assertion and Reason are true, Reason is NOT the correct explanation of Assertion.

Click Here to Download Notes

Key Concept: Basic Concepts

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that a homogeneous mixture has a uniform composition, which is true as by definition homogeneous mixtures have uniform properties and components are evenly distributed. The reason describes characteristics of a heterogeneous mixture where components are visibly separate and not uniformly mixed, which is also true. However, the statement about heterogeneous mixtures is not explaining why homogeneous mixtures have uniform composition. Thus, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Click Here to Download Notes

Key Concept: Basic Definition

c) A binary mixture with the same composition in liquid and vapor phase
[Solution Description] An azeotrope is defined as a binary mixture that has the same composition in both its liquid and vapor phases and boils at a constant temperature. This unique property makes it difficult to separate using conventional distillation techniques.

Click Here to Download Notes

Key Concept: Colligative Properties, Abnormal Colligative Properties

c) 1.5 K

[Solution Description] Boiling point elevation is calculated using:

$$\Delta T_b = i \cdot K_b \cdot m$$

Given $$\Delta T_b = 1.5\, K$$ and $i = 0.8$, we find $m$:

$$1.5 = 0.8 \cdot K_b \cdot m \quad \Rightarrow \quad m = \frac{1.5}{0.8 \cdot K_b}$$

The change in freezing point $\Delta T_f$ is expressed as:

$$\Delta T_f = i \cdot K_f \cdot m$$

Substituting for $m$ and since $K_f = K_b$:

$$\Delta T_f = 0.8 \cdot K_b \cdot \frac{1.5}{0.8 \cdot K_b} = 1.5 K$$

Therefore, the expected change in freezing point is 1.5 K.

Click Here to Download Notes