The number of moles of electrons passed when a current of 2 A is passed through a solution of an electrolyte for 20 minutes is:

4.1 \times 10^{- 4} m o l e^{-}

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1.24 \times 10^{- 2} m o l e^{-}

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2.487 \times 10^{- 2} m o l e^{-}

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2.487 \times 10^{- 1} m o l e^{-}

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Solution

The correct option is C $2.487 \times 10^{- 2} m o l e^{-}$
No.of moles of $e^{-} = \frac{Q}{F}$
$Q = I t$ where the time is in seconds.
$= \frac{2 \times 20 \times 60}{96500} = 2.487 \times 10^{- 2} m o l e^{-}$

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Q. The number of moles of electrons passed when a current of

2

A is passed through a solution of electrolyte for

20

minutes is:

4 m o l e s

of A are mixed with

4 m o l e s

of B when

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of C are formed at equilibrium. According to the equation

A + B ⇋ C + D

. Find the equilibrium constant:

Question 4
A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

(a) $10^{20}$
(b) $10^{16}$
(c) $10^{18}$
(d) $10^{23}$

Q. Question 4
A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

(a)

10^{20}

(b)

10^{16}

(c)

10^{18}

(d)

10^{23}

Q. A current of

1 A

is drawn by a filament of an electric bulb. What will be the approximate number of electrons passing through a cross section of its filament in 16 seconds?

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The number of moles of electrons passed when a current of 2 A is passed through a solution of an electrolyte for 20 minutes is:

The correct option is C 2.487×10−2 mol e−No.of moles of e− =QF Q=It where the time is in seconds. =2×20×6096500= 2.487×10−2 mol e−

The correct option is C $2.487 \times 10^{- 2} m o l e^{-}$
No.of moles of $e^{-} = \frac{Q}{F}$
$Q = I t$ where the time is in seconds.
$= \frac{2 \times 20 \times 60}{96500} = 2.487 \times 10^{- 2} m o l e^{-}$