Chemistry Test 241

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Chemistry Test 241

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Weekly Quiz Competition

Question 1
4 / -1

A solution of complex salt, CrCl₃.6H₂O having co-ordination no. 6 shows an Osmotic pressure 73.89 atm at 300 K, when 1.0 molar solution of complex is used. One litre of this solution on treatment with excess of AgNO₃ solution (α_AgNO3 =1) leads to the formation of AgCl :

A
1

B
2

C
3

D
1.5
Solution
Concept:
- When a pure solvent and solution are kept with a semi-permeable membrane between them then the solvent particles pass through the membrane from the solvent side to the solution side.
- This phenomenon is called “Osmosis”.
- The semi-permeable membrane is a membrane that allows only small molecules to pass through and blocks the larger solute molecules.
- The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis, i.e., to stop the passage of solvent molecules through a semipermeable membrane into the solution.
- Van’t Hoff Factor: To calculate the extent of association or dissociation, Van’t Hoff introduced factor i, known as the Van’t Hoff Factor.
Conclusion:

Thus, one liter of this solution on treatment with an excess of AgNO₃ solution (α_AgNO3 =1) leads to the formation of AgCl =2 moles.
Question 2
4 / -1

A current of 0.1A was passed for 4 hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for copper deposition. (Atomic weight of Cu = 63.5 g)

A
79%

B
39.5%

C
63.25%

D
63.5%
Solution
CONCEPT:

Current Efficiency in Electrolysis
- Current efficiency refers to the percentage of the electric current that contributes to the actual deposition of metal.
- In electrolysis, the amount of substance deposited is determined by Faraday's laws, with efficiency calculated by comparing theoretical and actual deposits
Question 3
4 / -1

Angular part of wave function for an orbital is sin θ cos θ sin θ where θ = angle from z-axis. Angular node(s) are

A
xy plane only

B
yz plane only

C
xy and xz only

D
xy . yz and zx only
Solution
CONCEPT:

Angular Nodes in Wave Functions
- Angular nodes are regions or planes where the probability of finding an electron is zero, determined by the form of the angular part of the wave function.
- In wave functions, terms involving ( sinθ), ( cosθ), and trigonometric functions of (ϕ) (angle in the xy-plane) helps identify the location of these nodes.
- Nodes occur where these functions equal zero, corresponding to specific planes in 3D space.
Question 4
4 / -1

Equal masses of O₂ and SO₂ gases were mixed in a vessel of 5 L capacity at 20°C. The total pressure exerted by the mixture was recorded to be 12 atm. The pressure exerted by O₂ in atmosphere is.

A
2

B
4

C
6

D
8
Solution
CONCEPT:

Partial Pressure in a Gas Mixture
- Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.
- The partial pressure of each gas is proportional to the number of moles of that gas in the mixture.
- If equal masses of two gases are mixed, the gas with the lighter molar mass will have a higher mole fraction, leading to a higher partial pressure.
Question 5
4 / -1

Manganese forms non - stoichiometric oxide having the general formula MnO_x. The value of x for a compound that contained 63.7% Mn, is

A
2.08

B
1.958

C
1.858

D
2.18
Solution
CONCEPT:

Non-Stoichiometric Compounds
- Non-stoichiometric compounds do not have a fixed ratio of elements, allowing for variable compositions.
- To determine the formula of such a compound, we can use the percentage composition to find the ratio of atoms of each element.
- In this problem, we are given the mass percentage of manganese and need to calculate the value of x in MnO_x.
Question 6
4 / -1

The difference between the heat change at constant pressure and that at constant volume for the combustion of 46 g of ethanol at 27°C is

A
– 4.59 kJ

B
+ 4.59 kJ

C
+ 2.49 kJ

D
+ 1.24 kJ
Solution
CONCEPT:

Difference Between Heat at Constant Pressure and Volume
- The heat change at constant pressure (ΔH) and at constant volume (ΔU) differ due to the work associated with the expansion or compression of gases.

Question 7

4 / -1

Match the columns

Column I (Solute in water)	Column II (van’t Hoff factor considering complete dissociation)
A. Glucose	i. 2.0
B. NaCl	ii. 1.0
C. MgCl₂	iii. 4.0
D. AlCl₃	iv. 3.0

A- i, B- ii, C- iii, D-iv

A- ii, B- i, C- iii, D-iv

A- ii, B- i, C- iv, D-iii

A- iv, B- ii, C- ii, D-iii

Solution

Concept:

The van’t Hoff factor (i) is used to express the effect of solute particles on the colligative properties of solutions. It varies based on the dissociation or association of the solute in water:

The van't Hoff factor measures the number of particles a solute splits into or forms in a solution.
It is particularly important in understanding colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure.
In case of dissociation, the van't Hoff factor is greater than 1.
In case of association, the van't Hoff factor is less than 1.

Explanation:

A. Glucose - i = 1.0 (Glucose does not dissociate in water, so the van’t Hoff factor is 1.0)
B. NaCl - i = 2.0 (NaCl dissociates into Na⁺ and Cl⁻, so the van’t Hoff factor is 2.0)
C. MgCl₂ - i = 3.0 (MgCl₂ dissociates into one Mg²⁺ and two Cl⁻ ions, so the van’t Hoff factor is 3.0)
D. AlCl₃ - i = 4.0 (AlCl₃ dissociates into one Al³⁺ and three Cl⁻ ions, so the van’t Hoff factor is 4.0)

Conclusion:

The correct matchup is: A- ii, B- i, C- iv, D-iii.

Question 8
4 / -1

The following data were obtained at a certain temperature for the decomposition of NH₃ in contact with catalyst.

The order of reaction is

A
2

B
3

C
1

D
None of these

Solution
Question 9
4 / -1

In van der Waals equation of state for a real gas, the term that accounts for the intermolecular forces is :

A
V_m – b

B
p + a/V_m²

C
RT

D
1/RT
Solution
CONCEPT:

Van der Waals Equation for Real Gases
- The ideal gas law, pV = nRT, assumes that gas molecules have no volume and do not interact with each other through intermolecular forces. However, real gases do not strictly follow this behavior, especially under high pressure and low temperature.
- To describe real gases more accurately, the van der Waals equation introduces two correction factors into the ideal gas law. The modified equation is:
- In this equation:
  
  p : The observed pressure of the gas.
  
  V_m : The molar volume of the gas, or the volume occupied by one mole of gas molecules.
  
  R : The gas constant.
  
  T : The absolute temperature of the gas.
  
  a : A constant specific to each gas that accounts for intermolecular attractions.
  
  b : A constant specific to each gas that accounts for the finite size of gas molecules.
EXPLANATION:
Question 10
4 / -1

In aqueous solution, which of the following orders regarding acid strength is correct?

A
HCN < CH₃COOH < HCl < HClO₄

B
HCN < CH₃COOH < HClO₄ < HCl

C
HCN < HClO₄ < HCl < CH₃COOH

D
HClO₄ < CH₃COOH < HCl < HCN
Solution
CONCEPT:

Understanding Acid Strength in Aqueous Solution
- Acid strength is defined by an acid’s ability to donate protons (H⁺) when dissolved in water.
- In aqueous solution, a stronger acid dissociates more completely, releasing a higher concentration of H⁺ ions. This leads to a lower pH.
- Acid strength is often influenced by the stability of the conjugate base formed after the acid donates its proton. More stable conjugate bases correlate with stronger acids.
EXPLANATION:
- HCN (Hydrocyanic acid):
  
  HCN is a very weak acid because its conjugate base, CN^-, is only moderately stabilized in solution.
  
  It only slightly dissociates in water, leading to a low concentration of H⁺ ions.
- CH₃COOH (Acetic acid):
  
  CH₃COOH is a weak acid, stronger than HCN, but it still only partially dissociates in water.
  
  The conjugate base CH₃COO^- is somewhat stabilized by resonance, allowing CH₃COOH to release H⁺ ions more easily than HCN.
- HCl (Hydrochloric acid):
  
  HCl is a strong acid and fully dissociates in aqueous solution, releasing a high concentration of H⁺ ions.
  
  Its conjugate base, Cl^-, is highly stable in water, making HCl a strong proton donor.
- HClO₄ (Perchloric acid):
  
  HClO₄ is among the strongest acids because it dissociates completely in water.
  
  The conjugate base, ClO₄^-, is exceptionally stable due to resonance and charge distribution, allowing HClO₄ to donate protons very readily.
CONCLUSION:

The correct option is: Option 1) HCN < CH₃COOH < HCl < HClO₄

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Chemistry Test 241

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Chemistry Test 241

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SHARING IS CARING

Question 1

1

2

3

1.5

Solution

Question 2

79%

39.5%

63.25%

63.5%

Solution

Question 3

xy plane only

yz plane only

xy and xz only

xy . yz and zx only

Solution

Question 4

2

4

6

8

Solution

Question 5

2.08

1.958

1.858

2.18

Solution

Question 6

– 4.59 kJ

+ 4.59 kJ

+ 2.49 kJ

+ 1.24 kJ

Solution

Question 7

A- i, B- ii, C- iii, D-iv

A- ii, B- i, C- iii, D-iv

A- ii, B- i, C- iv, D-iii

A- iv, B- ii, C- ii, D-iii

Solution

Concept:

Explanation:

​Conclusion:

Question 8

2

3

1

None of these

Solution

Question 9

Vm – b

p + a/Vm2

RT

1/RT

Solution

Question 10

HCN < CH3COOH < HCl < HClO4

HCN < CH3COOH < HClO4 < HCl

HCN < HClO4 < HCl < CH3COOH

HClO4 < CH3COOH < HCl < HCN

Solution

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Conclusion:

V_m – b

p + a/V_m²

HCN < CH₃COOH < HCl < HClO₄

HCN < CH₃COOH < HClO₄ < HCl

HCN < HClO₄ < HCl < CH₃COOH

HClO₄ < CH₃COOH < HCl < HCN