What is the time (in sec) requried for depositing all the silver present in $125 m L$ of $1 M$ $A g {N O}_{3}$ solution by passing a current of $241.24 A$ ?
( $1 F = 96500$ coulombs)

10

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50

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100

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90

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Solution

The correct option is B $50$
Weight deposited $\Rightarrow W = \frac{I t E}{F}$

$1 \times 125 \times 10^{- 3} \times 108 = \frac{241.24 \times t \times 108}{96500 \times 1}$

$t = 50 s e c$

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Similar questions

What is the time (in sec) required for deposition all the silver present in 125ml of 1M $A g N O_{3}$ solution by passing a current of 241.25 amperes?

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What is the time (in sec) requried for depositing all the silver present in 125mL of 1M AgNO3 solution by passing a current of 241.24A? (1F=96500 coulombs)

The correct option is B 50Weight deposited ⇒W=ItEF1×125×10−3×108=241.24×t×10896500×1t=50 sec

What is the time (in sec) requried for depositing all the silver present in $125 m L$ of $1 M$ $A g {N O}_{3}$ solution by passing a current of $241.24 A$ ?
( $1 F = 96500$ coulombs)

The correct option is B $50$
Weight deposited $\Rightarrow W = \frac{I t E}{F}$

$1 \times 125 \times 10^{- 3} \times 108 = \frac{241.24 \times t \times 108}{96500 \times 1}$

$t = 50 s e c$