The amount of metals deposited when 965 C of electricity is passed through NaCl and AgNO $_{3}$ solutions are:

23 g and 108 g

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11.5 g and 54 g

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0.23 g and 1.08 g

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2.3 g and 10.8 g

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Solution

The correct option is B 0.23 g and 1.08 g
$W = \frac{M \times C \times t}{n \times 96, 500}$

$W_{N a} = \frac{23 \times 965}{1 \times 96, 500} g, W_{A g} = \frac{108 \times 965}{1 \times 96, 500} g$

$W_{N a} = 0.23 g, W_{A g} = 1.08 g$

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The amount of metals deposited when 965 C of electricity is passed through NaCl and AgNO3 solutions are:

The correct option is B 0.23 g and 1.08 gW=M×C×tn×96,500WNa=23×9651×96,500g, WAg=108×9651×96,500gWNa=0.23g, WAg=1.08g

The amount of metals deposited when 965 C of electricity is passed through NaCl and AgNO $_{3}$ solutions are:

The correct option is B 0.23 g and 1.08 g
$W = \frac{M \times C \times t}{n \times 96, 500}$

$W_{N a} = \frac{23 \times 965}{1 \times 96, 500} g, W_{A g} = \frac{108 \times 965}{1 \times 96, 500} g$

$W_{N a} = 0.23 g, W_{A g} = 1.08 g$