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Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively are
  • a)
    3, 1 and 2
  • b)
    1, 3 and 2 
  • c)
    3, 1 and 3
  • d)
    2, 3 and 2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic w...

If x = 1 ⇒ y = 3, z = 2
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Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic w...
Given:
Atomic weight of A = 7
Atomic weight of B = 27
Atomic weight of C = 48
Weight of A deposited = 2.1 g
Weight of B deposited = 2.7 g
Weight of C deposited = 7.2 g

To find: Valencies of A, B and C

Calculation:
1. Find the number of equivalents of A, B and C deposited using the formula:

Number of equivalents = Weight of substance deposited / Equivalent weight

where Equivalent weight = Atomic weight / Valency

Number of equivalents of A = 2.1 / (7/Valency of A)
Number of equivalents of B = 2.7 / (27/Valency of B)
Number of equivalents of C = 7.2 / (48/Valency of C)

2. As the electrolysis was performed under identical conditions, the number of equivalents of each substance should be the same. Therefore, equate the above equations:

2.1 / (7/Valency of A) = 2.7 / (27/Valency of B) = 7.2 / (48/Valency of C)

3. Simplify the equation:

Valency of A : Valency of B : Valency of C = 1 : 3 : 2

4. Therefore, the valencies of A, B and C are 1, 3 and 2 respectively.

Answer: Option B (1, 3 and 2)
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Community Answer
Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic w...
Using Faraday's second law
mA/EA=KB/EB=mC/EC
EA=MA/VA=7/VA. where V is valances
EB=MB/VB=27/VB.
EC=MC/VC=48/VC
using Faraday's second law
mA/EA=KB/EB=mC/EC
2.1g/(7/VA)=2.7g/(27/VB)=7.2g/(48/VC)
2.1g÷VA/(7)=2.7g×VB/(27)=7.2g×VC/(48)
0.3VA=0.1VB=1.5VC
choice "B" makes the above equation right.
0.3(1)=0.1(3)=1.5(2)
0.3=0.3=0.3. so, VA=1. VB=3. VC=2
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Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively area)3, 1 and 2b)1, 3 and 2c)3, 1 and 3d)2, 3 and 2Correct answer is option 'B'. Can you explain this answer?
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Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively area)3, 1 and 2b)1, 3 and 2c)3, 1 and 3d)2, 3 and 2Correct answer is option 'B'. 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 Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively area)3, 1 and 2b)1, 3 and 2c)3, 1 and 3d)2, 3 and 2Correct answer is option 'B'. 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 Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively area)3, 1 and 2b)1, 3 and 2c)3, 1 and 3d)2, 3 and 2Correct answer is option 'B'. Can you explain this answer?.
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