Q124 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Chemistry
Calculate the time required in second to deposit 5.85 g Ni from $\text{NiCl}_2$ solution by passing 10 A current (molar mass of Ni $= 58.5\ \text{g mol}^{-1}$)
- A$19.3$
- B$193$
- Correct answerC$1930$
- D$19300$
Solution
The reaction for the deposition of Ni is:
$\text{Ni}^{2+} + 2e^{-} \rightarrow \text{Ni}$
Number of moles of Ni deposited $= \dfrac{5.85}{58.5} = 0.1 \text{ mol}$
Since $1 \text{ mole}$ of Ni requires $2 \text{ moles}$ of electrons, the total charge required is:
$Q = 0.1 \times 2 \times 96500 = 19300 \text{ C}$
Using the formula $Q = I \times t$:
$19300 = 10 \times t$
$t = 1930 \text{ s}$
Answer: $1930$
$\text{Ni}^{2+} + 2e^{-} \rightarrow \text{Ni}$
Number of moles of Ni deposited $= \dfrac{5.85}{58.5} = 0.1 \text{ mol}$
Since $1 \text{ mole}$ of Ni requires $2 \text{ moles}$ of electrons, the total charge required is:
$Q = 0.1 \times 2 \times 96500 = 19300 \text{ C}$
Using the formula $Q = I \times t$:
$19300 = 10 \times t$
$t = 1930 \text{ s}$
Answer: $1930$
Hence, the correct answer is option: 'C'.