On passing $3 A$ of electricity for $50$ min $1.8 g$ metal deposits. The equivalent mass of metal is:

9.3

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19.3

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38.3

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39.5

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Solution

The correct option is A $19.3$
$W_{m e t a l} = \frac{E i t}{96500}$

$= \frac{E \times 3 \times 50 \times 60}{96500}$

$E = \frac{96500 \times w}{3 \times 50 \times 60} = \frac{96500 \times 1.8}{3 \times 50 \times 60} = 19.3$

Hence, option B is correct.

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Similar questions

On passing 3A of electricity for 50 min, 1.8 g metal deposits. The equivalent mass of metal is

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Q. Electrochemical equivalent is the amount of metal deposited when __________ coulombs of electricity is passed.

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On passing 3 A of electricity for 50 min 1.8 g metal deposits. The equivalent mass of metal is:

The correct option is A 19.3Wmetal=Eit96500 =E×3×50×6096500E=96500×w3×50×60=96500×1.83×50×60=19.3Hence, option B is correct.

On passing $3 A$ of electricity for $50$ min $1.8 g$ metal deposits. The equivalent mass of metal is:

The correct option is A $19.3$
$W_{m e t a l} = \frac{E i t}{96500}$

$= \frac{E \times 3 \times 50 \times 60}{96500}$

$E = \frac{96500 \times w}{3 \times 50 \times 60} = \frac{96500 \times 1.8}{3 \times 50 \times 60} = 19.3$

Hence, option B is correct.