When same quantity of electricity is passed through AgNO3 and H2SO4 solution connected in series 5-04 - 10-2 g of H2 is liberated- Calculate the mass of silver deposited-

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Standard XII

Chemistry

Faraday's Second Law

When same qua...

Question

When same quantity of electricity is passed through $A g N O_{3}$ and $H_{2} S O_{4}$ solution connected in series 5.04 $\times 10^{- 2}$ g of $H_{2}$ is liberated. Calculate the mass of silver deposited:

54 g

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0.54 g

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5.4 g

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10.8 g

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Solution

The correct option is C $5.4 g$
The equivalent weight of Ag $= 108.$
The equivalent weight of hydrogen $= 1.$
Mass of hydrogen liberated $= 5.04 \times 10^{- 2} g .$

Using second law of electrolysis:

$\frac{M a s s o f A g d e p o s i t e d}{M a s s o f h y d r o g e n l i b e r a t e d} = \frac{E q . W t o f A g}{E q . W t . o f H_{2}}$
Hence,

$\frac{M a s s o f A g d e p o s i t e d}{5.04 \times 10^{- 2}} = \frac{108}{1}$
Mass of Ag deposited $= 5.04 \times 10^{- 2} \times 108 = 5.4 g .$

Option C is correct.

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When same quantity of electricity is passed through AgNO3 and H2SO4 solution connected in series 5.04×10−2 g of H2 is liberated. Calculate the mass of silver deposited:

When same quantity of electricity is passed through $A g N O_{3}$ and $H_{2} S O_{4}$ solution connected in series 5.04 $\times 10^{- 2}$ g of $H_{2}$ is liberated. Calculate the mass of silver deposited: