Passage of three faradays of charge through an aqueous solution of $A g N O_{3}, C u S O_{4}, A l (N O_{3})_{3}$ and $N a C l$ will deposit metals at the cathode in the molar ratio of:

1 : 2 : 3 : 1

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6 : 3 : 2 : 6

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6 : 3 : 0 : 0

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3 : 2 : 1 : 0

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Solution

The correct option is D $6 : 3 : 2 : 6$
$A g N O_{3} + e^{-} ⟶ A g + N O_{3}^{-}$
$1$ mole of $A g N O_{3}$ required $1$ mole of $e^{-}$ or $1$ Faraday of electricity.
Thus $3 F$ is used for $3$ mole of $A g N O_{3}$ .
$C u S O_{4} + 2 e^{-} ⟶ C u + S O_{4}^{2 -}$
$1$ mole of $C u S O_{4}$ required $2$ Faraday of charge.
Thus $3$ Faraday of charge will deposit $\frac{3}{2}$ mole of $C u S O_{4}$ .
$A l (M O_{3})_{3} + 3 e^{-}$ & $A l + 3 N O_{3}^{-}$
$3 F$ of charge is required for $1$ mole of $A l (N O_{3})_{3}$ .

$N a C l + e^{-} ⟶ N a + C l^{-}$
$1 F$ of charge required for $1$ mole of $N a C l$ .
Thus, $3 F$ of charge is required for $3$ mole of $N a C l$ .
The molar ratio will be:
$3 : \frac{3}{2} : 1 : 3$
or, $6 : 3 : 2 : 6$ .

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Passage of three faradays of charge through an aqueous solution of AgNO3,CuSO4,Al(NO3)3 and NaCl will deposit metals at the cathode in the molar ratio of:

Passage of three faradays of charge through an aqueous solution of $A g N O_{3}, C u S O_{4}, A l (N O_{3})_{3}$ and $N a C l$ will deposit metals at the cathode in the molar ratio of: