The density of Cu is 8-94g cm -3- The quantity of electricity needed to plate an area 10 cm - 10 cm to a thickness of 10-2 cm using CuSO4 solution would be-

The correct option is B 27172 CMass of copper deposited =density × volume = 8.94× 1= 8.94 gm.So according to faraday's law: W=ZQ=E96500× Q #160; ...

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Standard XII

Chemistry

Faraday's First Law

The density o...

Question

The density of Cu is 8.94g cm $^{- 3}$ . The quantity of electricity needed to plate an area $10 c m \times 10 c m$ to a thickness of $10^{- 2} c m$ using $C u S O_{4}$ solution would be:

13586 C

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27172 C

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40758 C

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20348 C

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Solution

The correct option is B 27172 C
Mass of copper deposited $= d e n s i t y \times v o l u m e = 8.94 \times 1 = 8.94$ gm.

So according to faraday's law:

$W = Z Q = \frac{E}{96500} \times Q$

$⟹ 8.94 = \frac{63.5 \times Q}{2 \times 96500}$

So, $Q = 27172$ Coulomb.

Option B is correct.

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The density of Cu is 8.94g cm−3. The quantity of electricity needed to plate an area 10cm×10cm to a thickness of 10−2cm using CuSO4 solution would be:

The correct option is B 27172 CMass of copper deposited =density×volume=8.94×1=8.94 gm.So according to faraday's law:W=ZQ=E96500×Q ⟹8.94=63.5×Q2×96500So, Q=27172 Coulomb.Option B is correct.

The density of Cu is 8.94g cm $^{- 3}$ . The quantity of electricity needed to plate an area $10 c m \times 10 c m$ to a thickness of $10^{- 2} c m$ using $C u S O_{4}$ solution would be:

The correct option is B 27172 C
Mass of copper deposited $= d e n s i t y \times v o l u m e = 8.94 \times 1 = 8.94$ gm.

So according to faraday's law:

$W = Z Q = \frac{E}{96500} \times Q$

$⟹ 8.94 = \frac{63.5 \times Q}{2 \times 96500}$

So, $Q = 27172$ Coulomb.

Option B is correct.