The ratio of weight of hydrogen and magnesium deposited by the same amount of electricity from $H_{2} {S O}_{4}$ and ${M g S O}_{4}$ in aqueous solution are:

1 : 8

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1 : 12

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1 : 16

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none of these

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Solution

The correct option is D none of these
According to Faraday's second law of electrolysis,
$\frac{W_{1}}{W_{2}} = \frac{E_{1}}{E_{2}}$
whereas, w is the weight of the substance deposited (in grams) while E is its chemical equivalent weight in gms per equivalent.
As we know that,
Equivalent weight $= \frac{w t .}{f}$ ; whereas, f = valence factor
Valence factor of $H_{2} S O_{4} = 2$
Mol. wt. of $H_{2} S O_{4} = 98 g$
Valence factor of $M g S O_{4} = 2$
Mol. wt. of $M g S O_{4} = 120 g$
$∴$ Eq. wt. of $H_{2} S O_{4} = \frac{98}{2} = 49$
Eq. wt. of $M g S O_{4} = \frac{120}{2} = 60$
$∴ \frac{W_{H_{2} S O_{4}}}{W_{M g S O_{4}}} = \frac{49}{60}$
$\Rightarrow W_{H_{2} S O_{4}} : W_{M g S O_{4}} = 49 : 60$

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