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If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would be
- a)88.8
- b)14.2
- c)9.0
- d)4.7
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
If a direct current deposits 20.5 g of potassium (atomic weight 39 u)...
At cathode; 
Equivalent weights are EK = 39/1 = 39 and EA1 = 27/3 = 9
From faraday’s second law of electrolysis,
So, 
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Community Answer
If a direct current deposits 20.5 g of potassium (atomic weight 39 u)...
To determine the amount of aluminium deposited by the same current in the given time interval, we can use Faraday's laws of electrolysis. According to Faraday's first law, the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Let's calculate the quantity of electricity passed through the electrolyte during the given time interval.
Given:
Mass of potassium deposited = 20.5 g
Atomic weight of potassium (K) = 39 u
We can use the formula:
Q = (m/M) * (Z/F)
where Q is the quantity of electricity (in coulombs), m is the mass deposited (in grams), M is the molar mass (in grams per mole), Z is the number of electrons involved in the reaction, and F is Faraday's constant (96485 C/mol).
For potassium:
m = 20.5 g
M = 39 g/mol
Z = 1 (as potassium has a valency of +1)
F = 96485 C/mol
Plugging in the values:
Q(K) = (20.5/39) * (1/96485)
Q(K) ≈ 5.3 * 10^-5 C
Now, let's calculate the mass of aluminium deposited using the same quantity of electricity.
Given:
Atomic weight of aluminium (Al) = 27 u
Using the same formula:
Q(Al) = Q(K)
m(Al) = (Q(Al) * M(Al)) / (Z * F)
where m(Al) is the mass of aluminium deposited (in grams).
Plugging in the values:
m(Al) = (5.3 * 10^-5 * 27) / (1 * 96485)
m(Al) ≈ 4.7 * 10^-5 g
Therefore, the amount of aluminium deposited by the same current during the given time interval is approximately 4.7 * 10^-5 grams, which is option 'D'.
Let's calculate the quantity of electricity passed through the electrolyte during the given time interval.
Given:
Mass of potassium deposited = 20.5 g
Atomic weight of potassium (K) = 39 u
We can use the formula:
Q = (m/M) * (Z/F)
where Q is the quantity of electricity (in coulombs), m is the mass deposited (in grams), M is the molar mass (in grams per mole), Z is the number of electrons involved in the reaction, and F is Faraday's constant (96485 C/mol).
For potassium:
m = 20.5 g
M = 39 g/mol
Z = 1 (as potassium has a valency of +1)
F = 96485 C/mol
Plugging in the values:
Q(K) = (20.5/39) * (1/96485)
Q(K) ≈ 5.3 * 10^-5 C
Now, let's calculate the mass of aluminium deposited using the same quantity of electricity.
Given:
Atomic weight of aluminium (Al) = 27 u
Using the same formula:
Q(Al) = Q(K)
m(Al) = (Q(Al) * M(Al)) / (Z * F)
where m(Al) is the mass of aluminium deposited (in grams).
Plugging in the values:
m(Al) = (5.3 * 10^-5 * 27) / (1 * 96485)
m(Al) ≈ 4.7 * 10^-5 g
Therefore, the amount of aluminium deposited by the same current during the given time interval is approximately 4.7 * 10^-5 grams, which is option 'D'.
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If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer?
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If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer?.
If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a direct current deposits 20.5 g of potassium (atomic weight 39 u) in 1 min, then the amount (in grams) of aluminium (atomic weight 27 u) deposited by the same current during the same time interval would bea)88.8b)14.2c)9.0d)4.7Correct answer is option 'D'. Can you explain this answer?.
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