Class 12 Chemistry Chapter 1 Solution

Key Concept: Component Identification

b) Solvent
[Solution Description] In this solid solution, palladium is present in the largest quantity compared to hydrogen, acting as the solvent and determining the physical state of the solution.

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Key Concept: Concentration Expression

b) Mass percentage
[Solution Description] One way to express the concentration of a solution is by using mass percentage, which is calculated as $$\left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100$$.

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Key Concept: Basic Definitions

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

[Solution Description]

The assertion describes the basic principle of osmosis correctly, where the solvent moves from a lower solute concentration to a higher solute concentration across a semipermeable membrane. This process occurs naturally and does not require energy.

The reason also accurately explains reverse osmosis, which is an applied version of osmosis. In reverse osmosis, external pressure is applied to force the solvent (often water) to move against its natural osmotic gradient, i.e., from high solute concentration to low solute concentration. This is commonly used for water purification.

Both statements are true, and the reason correctly explains the assertion as it demonstrates the concept of reversing the natural osmotic flow for practical applications like desalination.

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Key Concept: Real-World Application, Effect of Temperature and Pressure on Solubility

c) Detrimental to aquatic life
[Solution Description] Hot water discharges increase the temperature of river water, which reduces the solubility of oxygen. During summer, when temperatures are already elevated, this reduction can lead to lower levels of dissolved oxygen, potentially harming aquatic life by making it difficult for them to obtain necessary oxygen.

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Key Concept: Solubility and Temperature

b) Decreases
[Solution Description] According to the syllabus, the solubility of gases in liquids decreases with an increase in temperature. This occurs because higher temperatures provide more kinetic energy to gas molecules, making it easier for them to escape from the liquid phase.

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Key Concept: Real World Application, Raoult’s Law Application

c) 18.26°C
[Solution Description] Freezing point depression is calculated using the formula $$\Delta T_f = i \cdot m \cdot K_f$$. For ethanol-water mixture, we assume $i = 1$ since ethanol does not ionize in solution.

To find molality (m), assume a representative mass of solvent (water):

– Let 1000 g of water represent 55.56 mol (since molar mass of water is about 18 g/mol).
– Mole fraction of ethanol, $$X_{\text{ethanol}} = 0.15$$

Find moles of ethanol, $$n = X_{\text{ethanol}} \times \text{total moles} = 0.15 \times (55.56 + n)$$

Simplifying: $$n = \frac{0.15 \times 55.56}{0.85} = 9.82$$ moles

Molality (m) is therefore $$m = \frac{9.82}{1}$$ (since 1 kg of water):

Calculate $\Delta T_f$:
$$
\Delta T_f = 1 \cdot 9.82 \cdot 1.86 = 18.26^\circ C
$$

Thus, the freezing point is depressed by approximately 18.26°C below the normal freezing point of pure water.

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Key Concept: Temperature Effect

b) Solubility increases
[Solution Description] For an endothermic dissolution process, increasing the temperature usually increases solubility since the added heat favors the forward dissolution reaction.

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Key Concept: Complex Raoult’s Law Problems, Ideal vs Non-Ideal Solutions

b) 51.90 mmHg
[Solution Description]

The molar mass of ethanol $(C_2H_5OH)$ is approximately 46 g/mol and that of water $(H_2O)$ is 18 g/mol.

First, we calculate the number of moles in each component:
– Moles of ethanol: $$\frac{100 \text{ g}}{46 \text{ g/mol}} = 2.17 \text{ mol}$$
– Moles of water: $$\frac{50 \text{ g}}{18 \text{ g/mol}} = 2.78 \text{ mol}$$

Next, we find the mole fraction of each component:
– Mole fraction of ethanol $(X_{\text{ethanol}})$: $\frac{2.17}{2.17 + 2.78} = 0.438$
– Mole fraction of water $(X_{\text{water}})$: $\frac{2.78}{2.17 + 2.78} = 0.562$

Using Raoult’s law to find partial pressures:
– Partial pressure of ethanol: $$P_{\text{ethanol}} = X_{\text{ethanol}} \times P^{\circ}_{\text{ethanol}} = 0.438 \times 80 = 35.04 \text{ mmHg}$$
– Partial pressure of water: $$P_{\text{water}} = X_{\text{water}} \times P^{\circ}_{\text{water}} = 0.562 \times 30 = 16.86 \text{ mmHg}$$

Total vapor pressure of the solution:
$$
P_{\text{total}} = P_{\text{ethanol}} + P_{\text{water}} = 35.04 + 16.86 = 51.90 \text{ mmHg}
$$

Therefore, the total vapor pressure of the solution is 51.90 mmHg.

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Key Concept: Concentration Calculations

c) Assertion is true, but Reason is false.

[Solution Description] Mass percentage and mole fraction are both methods for expressing the concentration of components in a solution. The assertion is true, as mass percentage provides a way to describe how much solute is present relative to the total solution weight. The reason is incorrect because while mole fraction can describe component proportions, it is not specifically used to express solubility; rather, it describes the ratio of moles of a specific component to the total moles in the mixture. Thus, the Assertion is true, but Reason is false.

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Key Concept: Comparison with Other Units

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

[Solution Description]

Both the assertion and reason are true because mass percentage and volume percentage are indeed different types of measurements. Mass percentage is calculated using the formula:

$$\text{Mass%} = \left( \frac{\text{Mass of solute}}{\text{Total mass of solution}} \right) \times 100$$

On the other hand, volume percentage would be based on the ratio of volumes rather than masses. Therefore, these two percentages measure concentration differently as implied by the reason statement.

Hence, the correct answer is (a).

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