Q124 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Chemistry
How long (in minutes) must a $2\ \text{A}$ current be passed through $\text{AgNO}_3$ solution to deposit $5.4\ \text{g Ag}$? [Molar mass of Ag $= 108\ \text{g mol}^{-1}$]
- Correct answerA$40.2\ \text{min}$.
- B$54.5\ \text{min}$.
- C$30.1\ \text{min}$.
- D$60.6\ \text{min}$.
Solution
According to Faraday's first law of electrolysis:
$m = \dfrac{E \times I \times t}{96500}$
Here, $m = 5.4\ \text{g}$, $I = 2\ \text{A}$, and equivalent weight of Ag, $E = \dfrac{108}{1} = 108\ \text{g eq}^{-1}$.
Substituting the values:
$5.4 = \dfrac{108 \times 2 \times t}{96500}$
$t = \dfrac{5.4 \times 96500}{108 \times 2}$
$t = 2412.5\ \text{s}$
Converting time into minutes:
$t = \dfrac{2412.5}{60}\ \text{min} \approx 40.2\ \text{min}$
Answer: $40.2\ \text{min}$.
$m = \dfrac{E \times I \times t}{96500}$
Here, $m = 5.4\ \text{g}$, $I = 2\ \text{A}$, and equivalent weight of Ag, $E = \dfrac{108}{1} = 108\ \text{g eq}^{-1}$.
Substituting the values:
$5.4 = \dfrac{108 \times 2 \times t}{96500}$
$t = \dfrac{5.4 \times 96500}{108 \times 2}$
$t = 2412.5\ \text{s}$
Converting time into minutes:
$t = \dfrac{2412.5}{60}\ \text{min} \approx 40.2\ \text{min}$
Answer: $40.2\ \text{min}$.
Hence, the correct answer is option: 'A'.