A constant current( $0.5$ amp) is passed for $1$ hour through (i) aqueous $A g N O_{3}$ , (ii) aqueous $C u S O_{4}$ and (iii) molten $A l F_{3}$ , separately. The ratio of the mass of the metals deposited on the cathode is? [ $M_{A g}, M_{C u}, M_{A l}$ are molar masses of the respective metals].

M_{A g} : 2 M_{C u} : 3 M_{A l}

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M_{A g} : M_{C u} : M_{A l}

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6 M_{A g} : 3 M_{C u} : 2 M_{A l}

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3 M_{A g} : 2 M_{C u} : M_{A l}

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Solution

The correct option is C $6 M_{A g} : 3 M_{C u} : 2 M_{A l}$
The ratio of the mass of metal deposited is directly proportional to their equivalent weights.

$A g N O_{3} ⟶ A g^{+} + {N O_{3}}^{-}$
Equivalent weight of $A g N O_{3} = M_{A g}$

$C u S O_{4} ⟶ C u^{2 +} + S O_{4}^{2 -}$
Equivalent weight of $C u S O_{4} = \frac{M_{C u}}{2}$

$A l F_{3} ⟶ A l^{3 +} + 3 F^{-}$
Equivalent weight of $A l F_{3} = \frac{M_{A l}}{3}$

So, the ratio of the mass of metal deposited on the cathode is: $6 M_{A g} : 3 M_{C u} : 2 M_{A l}$

Hence, option C is correct.

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