When $6 \times 10^{22}$ electrons are used in the electrolysis of a metalic salt, 1.9 gm of the metal is deposited at the cathode The atomic weight of that metal Is 57. So oxidation state of the metal in the salt is:

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Solution

The correct option is A 3
mol. of $e^{-} = \frac{6 \times 10^{22}}{6 \times 10^{23}} = 0.1$ mol
moles of metal $= \frac{1.9}{57} = 0.033$ mol
$m^{n +} + n e^{-} ⟶ m$
$∴$ if $0.1$ of $e^{-}$ is passed and $0.033$ mol of $M$ is deposited.
charge $= \frac{0.1}{0.033} = + 3$

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When 6×1022 electrons are used in the electrolysis of a metalic salt, 1.9 gm of the metal is deposited at the cathode The atomic weight of that metal Is 57. So oxidation state of the metal in the salt is:

The correct option is A 3mol. of e−=6×10226×1023=0.1 molmoles of metal =1.957=0.033 molmn++ne−⟶m∴ if 0.1 of e− is passed and 0.033 mol of M is deposited.charge =0.10.033=+3

When $6 \times 10^{22}$ electrons are used in the electrolysis of a metalic salt, 1.9 gm of the metal is deposited at the cathode The atomic weight of that metal Is 57. So oxidation state of the metal in the salt is:

The correct option is A 3
mol. of $e^{-} = \frac{6 \times 10^{22}}{6 \times 10^{23}} = 0.1$ mol
moles of metal $= \frac{1.9}{57} = 0.033$ mol
$m^{n +} + n e^{-} ⟶ m$
$∴$ if $0.1$ of $e^{-}$ is passed and $0.033$ mol of $M$ is deposited.
charge $= \frac{0.1}{0.033} = + 3$