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Comprehension Type
This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)
Passage I
1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecell of Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 V
Q.
Pure [Ag+] in the ore is
- a)8.4 x 10-6 M
- b)2.9 x 10-6 M
- c)7.0 x 10-11 M
- d)4.2 x 10-6 M
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Comprehension TypeThis section contains a passage describing theory, e...
Cell Reaction
∴ x = 8.4 x 10-6 M
[Ag+] = 8.4 x 10-6 mol L-1
= 8.4 x 10-6 x 0.350 mol in 350 mL
= 8.4 x 10-6 x 0.350 x 108 g in 350 mL
= 3.1752 x 10-4 g in 1.05 g sample
Most Upvoted Answer
Comprehension TypeThis section contains a passage describing theory, e...
Explanation:
Given information:
- Mass of lead ore = 1.05 g
- Volume of solution = 350 mL
- Ecell of Pt(H2) | H (1M) || Ag | Ag = 0.500 V
- EAg / Ag = 0.80 V
To determine the pure [Ag] in the ore, we need to use the Nernst equation, which relates the cell potential of an electrochemical cell to the concentrations of the species involved.
Nernst equation:
Ecell = E°cell - (0.0592/n)log([Ag+]/[Ag])
where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred
[Ag+] is the concentration of Ag+ ions
[Ag] is the concentration of Ag atoms
Step 1: Calculate the number of moles of Ag+ ions:
First, we need to calculate the number of moles of Ag+ ions present in the solution.
The molar mass of Ag is 107.87 g/mol.
Number of moles of Ag+ ions = mass of Ag / molar mass of Ag
= 1.05 g / 107.87 g/mol
≈ 9.73 × 10^-3 mol
Step 2: Calculate the concentration of Ag+ ions:
Next, we need to calculate the concentration of Ag+ ions in the solution.
Concentration of Ag+ ions = number of moles / volume of solution
= 9.73 × 10^-3 mol / 0.350 L
≈ 2.78 × 10^-2 M
Step 3: Use the Nernst equation:
Now, we can use the Nernst equation to calculate the concentration of Ag atoms.
Ecell = 0.500 V
E°cell = 0.80 V
n = 1 (since 1 electron is transferred in the half-cell reaction)
0.500 V = 0.80 V - (0.0592/1)log([Ag]/2.78 × 10^-2)
0.500 V = 0.80 V - 0.0592log([Ag]/2.78 × 10^-2)
-0.30 V = -0.0592log([Ag]/2.78 × 10^-2)
log([Ag]/2.78 × 10^-2) = 0.30 / 0.0592
log([Ag]/2.78 × 10^-2) ≈ 5.07
[Ag]/2.78 × 10^-2 ≈ 10^5.07
[Ag] ≈ (10^5.07)(2.78 × 10^-2)
[Ag] ≈ 8.44 × 10^-6 M
Therefore, the concentration of pure Ag in the ore is approximately 8.44 × 10^-6 M, which corresponds to option (a).
Given information:
- Mass of lead ore = 1.05 g
- Volume of solution = 350 mL
- Ecell of Pt(H2) | H (1M) || Ag | Ag = 0.500 V
- EAg / Ag = 0.80 V
To determine the pure [Ag] in the ore, we need to use the Nernst equation, which relates the cell potential of an electrochemical cell to the concentrations of the species involved.
Nernst equation:
Ecell = E°cell - (0.0592/n)log([Ag+]/[Ag])
where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred
[Ag+] is the concentration of Ag+ ions
[Ag] is the concentration of Ag atoms
Step 1: Calculate the number of moles of Ag+ ions:
First, we need to calculate the number of moles of Ag+ ions present in the solution.
The molar mass of Ag is 107.87 g/mol.
Number of moles of Ag+ ions = mass of Ag / molar mass of Ag
= 1.05 g / 107.87 g/mol
≈ 9.73 × 10^-3 mol
Step 2: Calculate the concentration of Ag+ ions:
Next, we need to calculate the concentration of Ag+ ions in the solution.
Concentration of Ag+ ions = number of moles / volume of solution
= 9.73 × 10^-3 mol / 0.350 L
≈ 2.78 × 10^-2 M
Step 3: Use the Nernst equation:
Now, we can use the Nernst equation to calculate the concentration of Ag atoms.
Ecell = 0.500 V
E°cell = 0.80 V
n = 1 (since 1 electron is transferred in the half-cell reaction)
0.500 V = 0.80 V - (0.0592/1)log([Ag]/2.78 × 10^-2)
0.500 V = 0.80 V - 0.0592log([Ag]/2.78 × 10^-2)
-0.30 V = -0.0592log([Ag]/2.78 × 10^-2)
log([Ag]/2.78 × 10^-2) = 0.30 / 0.0592
log([Ag]/2.78 × 10^-2) ≈ 5.07
[Ag]/2.78 × 10^-2 ≈ 10^5.07
[Ag] ≈ (10^5.07)(2.78 × 10^-2)
[Ag] ≈ 8.44 × 10^-6 M
Therefore, the concentration of pure Ag in the ore is approximately 8.44 × 10^-6 M, which corresponds to option (a).
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Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer?
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Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer?.
Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Comprehension TypeThis section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)Passage I1.05 g of lead ore containing impurity of Ag was dissolved in HNO3 and the volume was made 350 mL. A silver electrode was dipped in the solution and Ecellof Pt(H2) | H+ (1M)|| Ag+| Ag was 0.500 V at 298 K. E°Ag+/Ag = 0.80 VQ. Pure [Ag+] in the ore is a)8.4 x 10-6 Mb)2.9x 10-6 Mc)7.0x 10-11 Md)4.2x 10-6 MCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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