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NEET CHEMISTRYElectrochemistryEasy

Question

A steady current of 1.5 A flows through a copper voltmeter for 10 min. If the electrochemical equivalent of copper is 30×105 g C130 \times 10^{-5} \text{ g C}^{-1}, the mass of copper deposited on the electrode will be:

A

0.40 g

B

0.50 g

C

0.67 g

D

0.27 g

Step-by-Step Solution

According to Faraday's First Law of Electrolysis, the mass of a substance deposited at an electrode is given by the formula:

m=Z×Q=Z×I×tm = Z \times Q = Z \times I \times t

Where: Z=30×105 g C1Z = 30 \times 10^{-5} \text{ g C}^{-1} (electrochemical equivalent) I=1.5 AI = 1.5 \text{ A} (current) t=10 min=10×60 s=600 st = 10 \text{ min} = 10 \times 60 \text{ s} = 600 \text{ s} (time)

First, calculate the total charge (QQ): Q=I×t=1.5 A×600 s=900 CQ = I \times t = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ C}

Now, calculate the mass (mm) deposited: m=30×105 g C1×900 Cm = 30 \times 10^{-5} \text{ g C}^{-1} \times 900 \text{ C} m=27000×105 g=0.27 gm = 27000 \times 10^{-5} \text{ g} = 0.27 \text{ g}

Therefore, 0.27 g of copper is deposited on the electrode.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistrysteadycurrentthroughcoppervoltmeter

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