Chemistry Test 230

CONCEPT:

Determining Empirical Formula from Percent Composition

• Convert the percentage of each element to grams (assuming 100 g of the compound).
• Convert the grams to moles using the atomic masses.
• Find the simplest whole-number ratio of moles of each element.
• Write the empirical formula based on the simplest ratio.

Calculations:

• Percentage of A: 32%
• Percentage of B: 20%
• Percentage of C: 100% - 32% - 20% = 48%
• Atomic masses: A = 64 u, B = 40 u, C = 32 u

So, Empirical Formula ,Based on the simplest ratio, the empirical formula of the compound X is ABC₃

CONCEPT:

Quantum Numbers

• Principal Quantum Number (n)
  • Indicates the size and energy level of the orbital.
• Azimuthal Quantum Number (l)
  • Indicates the shape of the orbital.
• Magnetic Quantum Number (m_l)
  • Indicates the orientation of the orbital in space.
• Spin Quantum Number (m_s)
  • Indicates the orientation of the electron's spin.

• A. m_l → III. Orientation of orbital
• B. m_s → IV. Orientation of spin of electron
• C. l → I. Shape of orbital
• D. n → II. Size of orbital

Conclusion:-
The correct answer i A-III, B-IV, C-I, D-II

Concept:

In the periodic table, the melting points (mp) and boiling points (bp) of elements can vary across different groups. These trends are influenced by the type of bonding, atomic size, and structure of the elements.

Explanation:

Comparing the melting and boiling point trends down the groups of the periodic table:

• Group 1 (Alkali Metals): Generally, mp/bp decrease down the group.
• Group 2 (Alkaline Earth Metals): Variation is not straightforward, but generally mp tends to decrease slightly, while bp shows irregular trends.
• Group 17 (Halogens): Both mp/bp increase down the group due to increasing molecular size and stronger van der Waals forces.
• Group 13 (Boron Group): Show varied trends, usually not a straightforward increase or decrease.

Among these groups, the halogens (Group 17) show a clear trend where both melting points and boiling points increase down the group. This is primarily due to the increase in molecular size, which leads to stronger van der Waals forces that require more energy to overcome.

Conclusion:

Therefore, the correct group where melting points and boiling points increase down the group is: Group 17.

Explanation:-

Types of Bonds in Different Molecules

• Ethane (C₂H₆) (A):
  • Ethane consists of a single bond (σ-bond) between the two carbon atoms.
  • Therefore, there is one σ-bond between the carbon atoms.
  • 

Ethene (C₂H₄) (B):

• Ethene consists of a double bond between the two carbon atoms.
• This includes one σ-bond and one π-bond.

Carbon molecule, C₂ (C):

• The carbon molecule (C₂) has a double bond between the carbon atoms.
• In some representations, it is considered to have two π-bonds due to the different possible bonding structures of carbon.

• Ethyne (C₂H₂) (D):
  • Ethyne has a triple bond between the two carbon atoms.
  • This includes one σ-bond and two π-bonds.

Based on the above definitions, we can match the items from List I to List II:

• ethane (A) - one σ-bond (III)
• ethene (B) - one σ-bond and one π-bond (IV)
• carbon molecule, C₂ (C) - two π-bonds (II)
• ethyne (D) - one σ-bond and two π-bonds (I)

So the correct match is  A-III, B-IV, C-II, D-I

Explanation:-

2NO(g) ⇌ N₂(g) + O₂(g)

• [N₂] = 3.0 × 10^–3 M
• [O₂] = 4.2 × 10^–3 M
• [NO] = 2.8 × 10^–3 M
• Initial concentration of NO = 0.1 mol L^–1

Step 1: Calculate K_c

The equilibrium constant K_c is given by:

Conclusion:-

The correct degree of dissociation ( α ) of NO(g) at equilibrium is0.717.

CONCEPT:

Le Chatelier's Principle and Reaction Shifts

• Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
• Factors affecting the equilibrium include concentration, temperature, and pressure.

Explanation:-

• The given reaction is: 2SO₂(g) + O₂(g) \leftrightharpoons 2SO₃(g); \Delta H = -45.2 \text{ kcals}
• This reaction is exothermic, as indicated by the negative ΔH value.

• Option 1: Removing O₂ from the reaction mixture will shift the equilibrium to the left, decreasing the amount of SO₃.
• Option 2: Increasing the temperature of the system will shift the equilibrium to the left (endothermic direction) to absorb the added heat, decreasing the amount of SO₃.
• Option 3: Increasing the pressure of the system by adding an inert gas will not change the partial pressures of the reactants or products, so the equilibrium position will not shift.
• Option 4: Decreasing the volume of the reaction mixture will increase the pressure. Since there are fewer moles of gas on the product side (2 moles of SO₃) compared to the reactant side (3 moles of gases), the equilibrium will shift to the right, increasing the amount of SO₃.

The correct answer is option 4

CONCLUSION: The number of moles of SO₃ will increase when the volume of the reaction mixture is decreased, as this increases the pressure and shifts the equilibrium towards the side with fewer moles of gas (the product side).

CONCEPT:

Boiling Points of Group 16 Hydrides

• The boiling points of hydrides of Group 16 elements generally increase with increasing molecular mass.
• However, H₂O is an exception because it exhibits strong hydrogen bonding, which significantly raises its boiling point compared to other Group 16 hydrides.

Explanation:-

Statement I: The boiling point of hydrides of Group 16 elements follow the order H₂O > H₂Te > H₂Se > H₂S.

This statement is true. The boiling point of H₂O is highest due to strong hydrogen bonding. The boiling points of H₂Te, H₂Se, and H₂S follow the expected trend based on molecular mass.

Statement II: On the basis of molecular mass, H₂O is expected to have a lower boiling point than the other members of the group, but due to the presence of extensive H-bonding in H₂O, it has a higher boiling point.

This statement is also true. Compared to H₂Te, H₂Se, and H₂S, water (H₂O) has a significantly higher boiling point due to hydrogen bonding, despite its lower molecular mass.

Conclusion:-

The correct answer is Both Statement I and Statement II are true

CONCEPT:

Entropy (S)

• Entropy is a measure of the disorder or randomness in a system.
• Processes that increase disorder or the number of available microstates will increase entropy.

Explanation:-

• A. A liquid evaporates to vapour
  • Entropy increases because the gas phase is more disordered than the liquid phase.
• B. Temperature of a crystalline solid lowered from 130 K to 0 K
  • Entropy decreases because lower temperature generally means lower kinetic energy and disorder for solids.
• C. 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
  • Entropy increases because gases (CO₂ and H₂O) are produced, leading to more disorder.
• D. Cl₂(g) → 2Cl(g)
  • Entropy increases because breaking a diatomic molecule into two atoms increases disorder.

Conclusion:-

The correct answer is A, C and D

Concept: 

Kohlrausch ‘s law:

• The molar conductivity of an electrolyte in its infinitely dilute solution is the sum of constituent ions' conductivity.
• Let A₂B₃ is an electrolyte and molar conductance is Λ⁰_A2B3 and at infinite dilution molar conductance of anion and cation is λ⁰_A3+ and λ⁰_B2- respectively. Then,

Λ⁰ _A2B3   =2 x  λ⁰_A3+  + 3 x  λ⁰_B2-

Explanation:

So, for the solution of CH₃COOH, K₂SO₄, and H₂SO₄ at infinite dilution:

Explanation:-

1. Colligative Properties:
   - Colligative properties (such as boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure lowering) depend on the number of solute particles in a solution, not their identity.
2. Colloidal Solutions:
   - In colloidal solutions, the solute particles are much larger than those in true solutions and typically consist of aggregates of many molecules, meaning there are fewer particles overall compared to a true solution of a non-electrolyte at the same concentration.
3. Non-Electrolyte Solutions:
   - In a solution of a non-electrolyte, each solute molecule typically dissolves independently, resulting in a higher number of solute particles compared to a colloidal solution of the same concentration.

Conclusion:
Since colligative properties are directly proportional to the number of solute particles, a colloidal solution will have fewer effective particles compared to a non-electrolyte solution of the same concentration. Therefore, the colligative property of a colloidal solution will be Lower