A current of 2.0 A is passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At. mass: 177 u). The oxidation states of the metal in metal salt is:

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Solution

The correct option is C 3
Weight= $\frac{E \times i \times t}{F} = \frac{M i t}{n - f a c t o r \times F}$ $| E = \frac{M}{n - f a c t o r}$
$\Rightarrow 22.2 = \frac{177 \times 5 \times 2 \times 60 \times 60}{n - f a c t o r \times 96500}$
$∴ n - f a c t o r = 2.97 ≃ 3$
Hence oxidation state of metal= $+ 3$

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A current of 2.0 A is passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At. mass: 177 u). The oxidation states of the metal in metal salt is:

The correct option is C 3Weight= E×i×tF=Mitn−factor×F |E=Mn−factor⇒22.2=177×5×2×60×60n−factor×96500∴n−factor=2.97≃3Hence oxidation state of metal= +3