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How many grams of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1 Faraday = 96,500 C; Atomic mass of Co = 59 u) [NEET Kar. 2013]
- a)0.66
- b)4.0
- c)20.0
- d)40.0
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
How many grams of cobalt metal will be deposited when a solution of co...
Applying,
Equivalent weight of cobalt (II) = 59/2
I = 10 A
Time (t) = 109 min = 109 × 60 sec
Substituting these values we get,
I = 10 A
Time (t) = 109 min = 109 × 60 sec
Substituting these values we get,
Most Upvoted Answer
How many grams of cobalt metal will be deposited when a solution of co...
Given:
Current = 10 A
Time = 109 minutes = 109 × 60 s = 6540 s
1 Faraday = 96,500 C
Atomic mass of Co = 59 u
To find:
The mass of cobalt metal deposited
Solution:
First, we need to calculate the charge (Q) passed through the electrolyte using the formula:
Q = I × t
where
I is the current in amperes (A)
t is the time in seconds (s)
Given:
Current (I) = 10 A
Time (t) = 6540 s
Substituting the values in the formula, we get:
Q = 10 A × 6540 s
Q = 65,400 C
We know that 1 Faraday is equal to 96,500 C.
Therefore, the number of Faradays passed (n) can be calculated as:
n = Q / 96,500
n = 65,400 C / 96,500 C
n = 0.678 Faradays
The number of moles of cobalt (Co) deposited can be calculated using the equation:
n = m / M
where
n is the number of moles
m is the mass of the substance in grams (g)
M is the molar mass of the substance in grams per mole (g/mol)
Given:
Molar mass of Co (M) = 59 g/mol
Number of Faradays passed (n) = 0.678 Faradays
Substituting the values in the equation, we get:
0.678 = m / 59
m = 0.678 × 59
m = 40.1 g
Therefore, the mass of cobalt metal deposited is 40.1 grams.
Current = 10 A
Time = 109 minutes = 109 × 60 s = 6540 s
1 Faraday = 96,500 C
Atomic mass of Co = 59 u
To find:
The mass of cobalt metal deposited
Solution:
First, we need to calculate the charge (Q) passed through the electrolyte using the formula:
Q = I × t
where
I is the current in amperes (A)
t is the time in seconds (s)
Given:
Current (I) = 10 A
Time (t) = 6540 s
Substituting the values in the formula, we get:
Q = 10 A × 6540 s
Q = 65,400 C
We know that 1 Faraday is equal to 96,500 C.
Therefore, the number of Faradays passed (n) can be calculated as:
n = Q / 96,500
n = 65,400 C / 96,500 C
n = 0.678 Faradays
The number of moles of cobalt (Co) deposited can be calculated using the equation:
n = m / M
where
n is the number of moles
m is the mass of the substance in grams (g)
M is the molar mass of the substance in grams per mole (g/mol)
Given:
Molar mass of Co (M) = 59 g/mol
Number of Faradays passed (n) = 0.678 Faradays
Substituting the values in the equation, we get:
0.678 = m / 59
m = 0.678 × 59
m = 40.1 g
Therefore, the mass of cobalt metal deposited is 40.1 grams.
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How many grams of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1 Faraday = 96,500 C; Atomic mass of Co = 59 u) [NEET Kar. 2013]a)0.66b)4.0c)20.0d)40.0Correct answer is option 'C'. Can you explain this answer?
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