Grams of cobalt metal that will be deposited when a solution of $c o b a l t (I I) c h l o r i d e$ is electrolyzed with a current of $10 a m p e r e s$ for $109 m i n u t e s$ :
( $1 F a r a d a y = 96, 500 C$ ; Atomic mass of $C o = 59 u$ )

4.0

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20.0

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40.0

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0.66

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Solution

The correct option is B $20.0$
Using formula from the 1st law of Faraday:
$w = \frac{I \times t \times E}{96500}$

Where, $I = 10 A m p e r e s$
$t = 109 m i n u t e s = 109 \times 60 = 6540 s e c o n d s$
$E$ = equivalent weight of cobalt = $\frac{59}{2}$
Putting all the values in the above formula:
$w = \frac{10 \times 6540 \times 59}{96500 \times 2}$

= $19.99 = 20.0 g$ (approximately)

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Grams of cobalt metal that will be deposited when a solution of cobalt(II) chloride is electrolyzed with a current of 10 amperes for 109 minutes: (1 Faraday=96,500 C; Atomic mass of Co=59 u)

The correct option is B 20.0Using formula from the 1st law of Faraday: w=I×t×E96500 Where, I=10 Amperes t=109 minutes=109×60=6540 seconds E = equivalent weight of cobalt = 592 Putting all the values in the above formula: w=10×6540×5996500×2 = 19.99=20.0 g (approximately)

Grams of cobalt metal that will be deposited when a solution of $c o b a l t (I I) c h l o r i d e$ is electrolyzed with a current of $10 a m p e r e s$ for $109 m i n u t e s$ :
( $1 F a r a d a y = 96, 500 C$ ; Atomic mass of $C o = 59 u$ )