One gram of metal ion M^{3+} is discharged by | vedclass

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Question

(D) The reaction for the discharge of the metal ion is: $M^{3+} + 3e^- ightarrow M$.Number of moles of electrons = $\frac{1.81 	imes 10^{23}}{6.022

Vedclass · Accepted Answer

None of these

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(D) The reaction for the discharge of the metal ion is: $M^{3+} + 3e^- ightarrow M$.Number of moles of electrons = $\frac{1.81 	imes 10^{23}}{6.022 	imes 10^{23}} \approx 0.3 \ mol$.According to the stoichiometry,$3 \ mol$ of electrons discharge $1 \ mol$ of $M^{3+}$.Therefore,$0.3 \ mol$ of electrons will discharge $\frac{0.3}{3} = 0.1 \ mol$ of metal $M$.We are given that the mass of the discharged metal is $1 \ g$.Since $0.1 \ mol$ of metal weighs $1 \ g$,the molar mass (atomic weight) of the metal is $\frac{1 \ g}{0.1 \ mol} = 10 \ g \ mol^{-1}$.Since $10 \ g \ mol^{-1}$ is not among the options,the correct choice is $D$.

One gram of metal ion $M^{3+}$ is discharged by the passage of $1.81 \times 10^{23}$ electrons. What is the atomic weight of the metal?

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