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Q.27. 9650 charges is passed through an aqueous solution of metal nitrate $M(N{O}_3{)}_X$, to obtain 2g metal (at. Mass 180). The valence of metal is

Answer 1

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$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{second}\mathrm{law}\mathrm{of}\mathrm{electrolysis}, \\ \frac{\mathrm{w}}{\mathrm{E}}=\frac{\mathrm{It}}{96500} \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{charge}=9650=\mathrm{It} \\ \mathrm{So}, \\ \frac{2}{\mathrm{E}}=\frac{9650}{96500} \\ \mathrm{or},\mathrm{E}=20 \\ \mathrm{E}=\frac{\mathrm{Atomic}\mathrm{weight}}{\mathrm{Valency}} \\ \mathrm{So},\frac{180}{\mathrm{Valency}}=20 \\ \mathrm{or},\mathrm{Valency}=9 \end{gathered}$$