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Calculate the mass of copper deposited if a current of 0.45A...

Calculate the mass of copper deposited if a current of 0.45A flows through $CuSO_{4}$ solution for 1hour 15mins. [Cu=64.0, S= 32.0, O =16.0, 1F= 96500C]

A.

6.40g

B.

0.67g

C.

0.64g

D.

0.45g

View answer & explanation

Correct answer is B

$Cu^{2+} + 2e^{-} \to Cu_{(s)}$

$2 \times 96500C = 193000C deposits 64g of Cu$

$M \propto It$

$Q = It = 0.45A \times 1hr15mins$

= $0.45 \times 75 \times 60$ [converting the time to seconds]

Quantity of electricity passed = 2025C

$193000C \to 64g Cu$

$1C \to \frac{64}{193000}$

$2025C \to \frac{64}{193000} \times 2025$

$= 0.6715g \approxeq 0.67g$

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