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1.0 g of chromium metal (molar mass = 52.00 g/mol) is deposited by the electrolysis of a Cr(NO3)3 solution. If a current of 3.0 A was used, what is the duration of the current in seconds?

A. 3700 seconds
B. 360 seconds
C. 1900 seconds
D. 630 seconds

Answer :

Final answer:

The duration of the current in seconds is approximately 1900 seconds.

Explanation:

To calculate the duration of the current in seconds, we can use Faraday's laws of electrolysis. According to Faraday's first law, the mass of a substance deposited is directly proportional to the amount of charge passed through the electrolyte.

The equation to calculate the amount of substance deposited is:

Mass (g) = (Current (A) × Time (s) × Molar mass (g/mol)) / (Faraday's constant (C/mol))

In this case, we have the mass of chromium metal (1.0 g), the current (3.0 A), and the molar mass of chromium (52.00 g/mol). We need to calculate the time (duration of the current) in seconds.

Let's plug in the values into the equation:

1.0 g = (3.0 A × Time (s) × 52.00 g/mol) / (Faraday's constant)

Now, we can rearrange the equation to solve for Time (s):

Time (s) = (1.0 g × Faraday's constant) / (3.0 A × 52.00 g/mol)

Faraday's constant is approximately 96,485 C/mol.

Let's substitute the values and calculate:

Time (s) = (1.0 g × 96,485 C/mol) / (3.0 A × 52.00 g/mol)

Time (s) ≈ 1900 seconds

Learn more about calculating the duration of current in electrolysis here:

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