$1$ mole of $A l$ is deposited by $X$ coulomb of electricity passing through aluminium nitrate solution. The number of mole of silver deposited by $X$ coulomb of electricity from silver nitrate solution is:

3

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Solution

The correct option is B $3$
${A l}^{3 +} + 3 e^{-} \to A l$
$X$ coloumb deposite $1$ mole of $A l$ $=$ $3$ eq. of $A l$
${A g}^{+} + e^{-} \to A g$
$∴$ Silver deposited $= 3$ eq $= 3$ mole

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mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution. Calculate the number of moles of silver deposited by X coulomb of electricity from silver nitrate solution.

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Q. When

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coulomb of electricity is passed through a solution of silver nitrate, the amount of silver deposited is:

When 9.65 coulombs of electricity is passed through a solution of silver nitrate

(atomic weight of taking as 108) the amount of silver deposited is

[EAMCET 1992; KCET 2000]

When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.3, approximately 108) the amount of silver deposited is:

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1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate solution. The number of mole of silver deposited by X coulomb of electricity from silver nitrate solution is:

The correct option is B 3Al3++3e−→AlX coloumb deposite 1 mole of Al = 3 eq. of AlAg++e−→Ag∴ Silver deposited =3 eq =3 mole

$1$ mole of $A l$ is deposited by $X$ coulomb of electricity passing through aluminium nitrate solution. The number of mole of silver deposited by $X$ coulomb of electricity from silver nitrate solution is:

The correct option is B $3$
${A l}^{3 +} + 3 e^{-} \to A l$
$X$ coloumb deposite $1$ mole of $A l$ $=$ $3$ eq. of $A l$
${A g}^{+} + e^{-} \to A g$
$∴$ Silver deposited $= 3$ eq $= 3$ mole