Chemistry Exam > Chemistry Questions > Salts of A (atomic weight 7), B (atomic weigh...
Start Learning for Free
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:
- 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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weigh...
According to faraday's law:
W = ZQ = EQ/96500
For A:
2.1=(7/x)Q/96500
For B:
2.7=(27/y)Q/96500
For C:
7.2=(48/z)Q/96500
x: y:z ::(7/2.1):(27/2.7):(48/7.2)
W = ZQ = EQ/96500
For A:
2.1=(7/x)Q/96500
For B:
2.7=(27/y)Q/96500
For C:
7.2=(48/z)Q/96500
x: y:z ::(7/2.1):(27/2.7):(48/7.2)
= 3.33:10:6.66
=1:3:2
by solving these equations,
x=1, y=3, z=2
by solving these equations,
x=1, y=3, z=2
Most Upvoted Answer
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weigh...
According to faraday's law:
W = ZQ = EQ/96500
For A:
2.1=(7/x)Q/96500
For B:
2.7=(27/y)Q/96500
For C:
7.2=(48/z)Q/96500
x: y:z ::(7/2.1):(27/2.7):(48/7.2)
W = ZQ = EQ/96500
For A:
2.1=(7/x)Q/96500
For B:
2.7=(27/y)Q/96500
For C:
7.2=(48/z)Q/96500
x: y:z ::(7/2.1):(27/2.7):(48/7.2)
= 3.33:10:6.66
=1:3:2
by solving these equations,
x=1, y=3, z=2
by solving these equations,
x=1, y=3, z=2
Free Test
FREE
| Start Free Test |
Community Answer
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weigh...
Given information:
- Salts of A, B, and C were electrolyzed under identical conditions using the same quantity of electricity.
- When 2.1 g of A was deposited, the weights of B and C deposited were 2.7 g and 7.2 g.
To find:
The valencies of A, B, and C.
Explanation:
The weight of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the solution and the number of moles of the substance being deposited.
We can calculate the number of moles of each substance deposited using their respective weights and atomic weights.
Calculations:
1. Number of moles of A deposited:
- Weight of A deposited = 2.1 g
- Atomic weight of A = 7 g/mol
- Number of moles of A = Weight of A deposited / Atomic weight of A = 2.1 g / 7 g/mol = 0.3 mol
2. Number of moles of B deposited:
- Weight of B deposited = 2.7 g
- Atomic weight of B = 27 g/mol
- Number of moles of B = Weight of B deposited / Atomic weight of B = 2.7 g / 27 g/mol = 0.1 mol
3. Number of moles of C deposited:
- Weight of C deposited = 7.2 g
- Atomic weight of C = 48 g/mol
- Number of moles of C = Weight of C deposited / Atomic weight of C = 7.2 g / 48 g/mol = 0.15 mol
Ratio of moles:
Now, let's calculate the ratio of moles of A, B, and C to find their valencies.
- A : B : C = 0.3 mol : 0.1 mol : 0.15 mol
Simplifying the ratio:
- A : B : C = 3 : 1 : 1.5
Since valencies are whole numbers, we can multiply the ratio by 2 to get the simplest whole number ratio:
- A : B : C = 6 : 2 : 3
Valencies:
From the simplified ratio, we can conclude that:
- The valency of A is 1 because it has a ratio of 6:2 with B.
- The valency of B is 3 because it has a ratio of 2:6 with A and C.
- The valency of C is 2 because it has a ratio of 3:6 with B.
Therefore, the valencies of A, B, and C are 1, 3, and 2 respectively, which corresponds to option B.
- Salts of A, B, and C were electrolyzed under identical conditions using the same quantity of electricity.
- When 2.1 g of A was deposited, the weights of B and C deposited were 2.7 g and 7.2 g.
To find:
The valencies of A, B, and C.
Explanation:
The weight of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the solution and the number of moles of the substance being deposited.
We can calculate the number of moles of each substance deposited using their respective weights and atomic weights.
Calculations:
1. Number of moles of A deposited:
- Weight of A deposited = 2.1 g
- Atomic weight of A = 7 g/mol
- Number of moles of A = Weight of A deposited / Atomic weight of A = 2.1 g / 7 g/mol = 0.3 mol
2. Number of moles of B deposited:
- Weight of B deposited = 2.7 g
- Atomic weight of B = 27 g/mol
- Number of moles of B = Weight of B deposited / Atomic weight of B = 2.7 g / 27 g/mol = 0.1 mol
3. Number of moles of C deposited:
- Weight of C deposited = 7.2 g
- Atomic weight of C = 48 g/mol
- Number of moles of C = Weight of C deposited / Atomic weight of C = 7.2 g / 48 g/mol = 0.15 mol
Ratio of moles:
Now, let's calculate the ratio of moles of A, B, and C to find their valencies.
- A : B : C = 0.3 mol : 0.1 mol : 0.15 mol
Simplifying the ratio:
- A : B : C = 3 : 1 : 1.5
Since valencies are whole numbers, we can multiply the ratio by 2 to get the simplest whole number ratio:
- A : B : C = 6 : 2 : 3
Valencies:
From the simplified ratio, we can conclude that:
- The valency of A is 1 because it has a ratio of 6:2 with B.
- The valency of B is 3 because it has a ratio of 2:6 with A and C.
- The valency of C is 2 because it has a ratio of 3:6 with B.
Therefore, the valencies of A, B, and C are 1, 3, and 2 respectively, which corresponds to option B.
|
Explore Courses for Chemistry exam
|
|
Similar Chemistry Doubts
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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?
Question Description
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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 Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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 Chemistry 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 electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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?.
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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 Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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 Chemistry 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 electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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?.
Solutions for Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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? in English & in Hindi are available as part of our courses for Chemistry.
Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Here you can find the meaning of Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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? defined & explained in the simplest way possible. Besides giving the explanation of
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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?, a detailed solution for Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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? has been provided alongside types of Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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? theory, EduRev gives you an
ample number of questions to practice Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition 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 g. The valencies of A, B and C respectively:a)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? tests, examples and also practice Chemistry tests.
|
|
Explore Courses for Chemistry exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup