A current of $2 a m p$ when passed for $5$ hours through a molten salt deposits $22.2 g$ of metal of atomic mass $177$ . The oxidation state of the metal in the metal salt is:

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Solution

The correct option is B $+ 3$
$W = \frac{I t E}{96500}$

$∴ E = \frac{W \times 96500}{I t} = \frac{22.2 \times 96500}{2 \times 5 \times 3600} = 59.5$

$∴ Equivalent mass = \frac{Atomic mass}{oxidation state}$

$59.5 = \frac{177}{n} ∴ n ≃ 3$

Hence, the correct option is $C$

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A current of 2 amp when passed for 5 hours through a molten salt deposits 22.2g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is:

The correct option is B +3W=ItE96500∴E=W×96500It=22.2×965002×5×3600=59.5∴Equivalent mass=Atomic mass oxidation state 59.5=177n∴n≃3Hence, the correct option is C

A current of $2 a m p$ when passed for $5$ hours through a molten salt deposits $22.2 g$ of metal of atomic mass $177$ . The oxidation state of the metal in the metal salt is:

The correct option is B $+ 3$
$W = \frac{I t E}{96500}$

$∴ E = \frac{W \times 96500}{I t} = \frac{22.2 \times 96500}{2 \times 5 \times 3600} = 59.5$

$∴ Equivalent mass = \frac{Atomic mass}{oxidation state}$

$59.5 = \frac{177}{n} ∴ n ≃ 3$

Hence, the correct option is $C$