Question

A current of 3 A is passed through a molten calcium salt for 1 hr 47 min 13 sec. The mass of calcium deposited is : (Molar mass of 1 Ca=40gmol^-1)

• A. 6.0 g
• B. 2.0 g
• C. 8.0 g
• D. 4.0 g

Answer 1

The Correct Option is D

To calculate the mass of calcium deposited, we can use Faraday's Law of Electrolysis, which states:

\[ \text{Mass} = \frac{M \times I \times t}{n \times F} \]

Where:

• M is the molar mass of calcium (40 g/mol),
• I is the current (3 A),
• t is the time in seconds,
• n is the number of electrons involved in the deposition of calcium (since calcium has a charge of +2, n = 2),
• F is the Faraday constant (96500 C/mol).

Step 1: Convert the time to seconds

The time is given as 1 hour 47 minutes 13 seconds.

1 hour = 3600 seconds, 47 minutes = 47 × 60 = 2820 seconds, 13 seconds = 13 seconds

Total time t:
t = 3600 + 2820 + 13 = 6433 seconds

Step 2: Apply Faraday's Law

\[ \text{Mass of Ca} = \frac{40 \, \text{g/mol} \times 3 \, \text{A} \times 6433 \, \text{sec}}{2 \times 96500 \, \text{C/mol}} \]

Simplifying the equation:

\[ \text{Mass of Ca} = \frac{40 \times 3 \times 6433}{2 \times 96500} \]

\[ \text{Mass of Ca} = \frac{772960}{193000} \]

\[ \text{Mass of Ca} = 4.0 \, \text{g} \]

Conclusion:

The mass of calcium deposited is 4.0 g, making option (D) the correct answer.

Answer 2

Using Faraday's law of electrolysis: \[ \text{Mass deposited} = \frac{M \times I \times t}{n \times F} \] Where:
\( M \) is the molar mass of calcium (40 g/mol),
\( I \) is the current (3 A),
\( t \) is the time in seconds (1 hr 47 min 13 sec = 6423 s),
\( n \) is the number of electrons involved in the reaction (for Ca, \( n = 2 \)),
\( F \) is Faraday's constant (96,485 C/mol).
Substitute the values: \[ \text{Mass} = \frac{40 \times 3 \times 6423}{2 \times 96,485} = 4.0 \, g \]