Number of electrons involved in the electrodeposition of $63.5 g$ of $C u$ from a solution of $C u S O_{4}$ is?

6.022 \times 10^{23}

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3.011 \times 10^{23}

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12.044 \times 10^{23}

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6.022 \times 10^{22}

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Solution

The correct option is C $12.044 \times 10^{23}$
$\begin{matrix} 63.5 g m o f C u = 1 m o l o f C u . N o . o f e q u i v a l e n t = N o . o f m o l e s \times v a l e n c y f a c t o r = 1 \times 2 = 2 1 m o l e o f e l e c t r o n w i l l d e p o s i t e 1 e q u i v a l e n t . N u m b e r o f e l e c t r o n = 2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23} . H e n c e, o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

One mole of any substance contains $6.022 \times 10^{23}$ atoms/molecules. Number of molecules of $H_{2} S O_{4}$ present in 100 mL of $0.02 M H_{2} S O_{4}$ solution is.............

(a) $12.044 \times 10^{20}$ molecules (b) $6.022 \times 10^{23}$ molecules

If 32 g of $O_{2}$ contains 6.022 $\times 10^{23}$ molecules at NTP then 32 g of S, under the same conditions, will contain,

Q. Number of electrons involved in the electrode position of

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is :

(N_{A} = 6 \times 10^{23}

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Question 9
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68 \times 10^{23}$
(b) $6.09 \times 10^{22}$
(c) $6.022 \times 10^{23}$
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Number of electrons involved in the electrodeposition of 63.5 g of Cu from a solution of CuSO4 is?

The correct option is C 12.044×102363.5gmofCu=1molofCu.No.ofequivalent=No.ofmoles×valencyfactor=1×2=21moleofelectronwilldeposite1equivalent.Numberofelectron=2×6.022×1023=12.044×1023.Hence,optionCisthecorrectanswer.

Number of electrons involved in the electrodeposition of $63.5 g$ of $C u$ from a solution of $C u S O_{4}$ is?