Q124 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Chemistry
Find the mass of Zn metal deposited at the cathode during the passage of $10$ amperes current through its salt solution for $10$ minutes. [ Molar mass of Zn $= 65.4$ g $\text{mol}^{-1}$ & For the reaction of Zn, mole ratio $= 1/2$ ]
- A$1.0$ g
- Correct answerB$2.0$ g
- C$3.0$ g
- D$4.0$ g
Solution
Given:
Current $I = 10$ A
Time $t = 10$ minutes $= 10 \times 60 = 600$ s
Molar mass of Zn $M = 65.4$ g mol$^{-1}$
Valency factor $n = 2$ (since mole ratio is $1/2$)
Faraday's constant $F = 96500$ C mol$^{-1}$
According to Faraday's first law of electrolysis:
$m = \dfrac{M}{nF}It$
Substituting the values:
$m = \dfrac{65.4}{2 \times 96500} \times 10 \times 600$
$m = \dfrac{65.4 \times 6000}{193000}$
$m = \dfrac{392.4}{193} \approx 2.03$ g
The closest value is $2.0$ g.
Answer: $2.0$ g
Current $I = 10$ A
Time $t = 10$ minutes $= 10 \times 60 = 600$ s
Molar mass of Zn $M = 65.4$ g mol$^{-1}$
Valency factor $n = 2$ (since mole ratio is $1/2$)
Faraday's constant $F = 96500$ C mol$^{-1}$
According to Faraday's first law of electrolysis:
$m = \dfrac{M}{nF}It$
Substituting the values:
$m = \dfrac{65.4}{2 \times 96500} \times 10 \times 600$
$m = \dfrac{65.4 \times 6000}{193000}$
$m = \dfrac{392.4}{193} \approx 2.03$ g
The closest value is $2.0$ g.
Answer: $2.0$ g
Hence, the correct answer is option: 'B'.