The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous $H_{2} S O_{4}$ and fused $M g S O_{4}$ are:

1:8

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1:12

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1:16

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1:1

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Solution

The correct option is A 1:12
Acc. to Faraday $2^{n d}$ Law of electrolysis.
$\frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} w_{1} =$ At deposited of element H.
$\frac{w_{1}}{w_{2}} = \frac{1}{12} w_{2} =$ At deposited of element Mg.
$E_{1} =$ eq. at of N.
$E_{2} =$ eq. at of Mg.

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The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous H2SO4 and fused MgSO4 are:

The correct option is A 1:12Acc. to Faraday 2nd Law of electrolysis.w1w2=E1E2w1= At deposited of element H. w1w2=112w2= At deposited of element Mg. E1= eq. at of N. E2= eq. at of Mg.

The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous $H_{2} S O_{4}$ and fused $M g S O_{4}$ are:

The correct option is A 1:12
Acc. to Faraday $2^{n d}$ Law of electrolysis.
$\frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} w_{1} =$ At deposited of element H.
$\frac{w_{1}}{w_{2}} = \frac{1}{12} w_{2} =$ At deposited of element Mg.
$E_{1} =$ eq. at of N.
$E_{2} =$ eq. at of Mg.