The amount of $A g$ (atomic mass $= 108$ ) deposited at the cathode when a current of $0.5 a m p$ is passed through a solution of $A g N O_{3}$ for $1$ hour is closest to

2 g

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5 g

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108 g

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11 g

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Solution

The correct option is B $2 g$
From the law of electrolysis,

$W = \frac{E}{96500} \times I \times t$

$W = \frac{108}{96500} \times 0.5 \times 3600 = 2 g m$ .

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Similar questions

The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO $_{3}$ for 1 hour is closest to :

Q. Calculate the mass of

A g

deposited at cathode when a current of 2 amperes was passed through a solution of

A g N O_{3}

for 15 minutes.

Q. Calculate the mass of Ag deposited at cathode when a current of 2A was passed through a solution of

A g {N O}_{3}

for 15 min.
(Given : Molar mass of

A g = 108 g {m o l}^{- 1}

1 F = 96500 C {m o l}^{- 1}

Q. Find the mass of silver deposited when a current of 1.5 ampere is passed through silver nitrate solution for 5 min.

(Atomic mass of

A g

108

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The amount of Ag (atomic mass =108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO3 for 1 hour is closest to

The correct option is B 2 gFrom the law of electrolysis,W=E96500×I×tW=10896500×0.5×3600=2 gm.

The amount of $A g$ (atomic mass $= 108$ ) deposited at the cathode when a current of $0.5 a m p$ is passed through a solution of $A g N O_{3}$ for $1$ hour is closest to

The correct option is B $2 g$
From the law of electrolysis,

$W = \frac{E}{96500} \times I \times t$

$W = \frac{108}{96500} \times 0.5 \times 3600 = 2 g m$ .